Sat u odlicnom condition. Payment details: Brand: swiss it repair Jaeger LeCoultre Series: Older Mo

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Sat u odlicnom condition. Payment details: Brand: swiss it repair Jaeger LeCoultre Series: Older Models Retail Price: $ 10150 Reference: Q1428420 As Known As: Q1428420 Case Size: 38mm x 10mm Gender: Gents Movement: Automatic Functions: Date, Second Time Zone, Power reserve indicator, Hours, Minutes, Seconds Strap: Leather Strap Signatures: Jaeger LeCoultre Description: Second time zone selector swiss it repair at 6, Date indicator at 2, Power reserve at 10 Water resistant to 165 feet ili pm Contact 0612137613 Price Jaeger LeCoultre Master Geographic 2700e.Pozdrav ....
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4 March 1931, the Reverso was patented 80 years of history are celebrated for the iconic reversible

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4 March 1931, the Reverso was patented 80 years of history are celebrated for the iconic reversible watch 2011 is the year of the Reverso watch, which celebrates 80 years of rich history, having adorned the wrists of polo players and illustrious personalities. world gym bayshore Yet, before we can begin to glorify these eight decades of watchmaking history, we need to go back to 4 March 1931, when it all began. Looking back to 4 March 1931, at 1:15 pm, at the INPI (National Industrial Property Institute) in Paris: René-Alfred Chauvot world gym bayshore officially patents his invention of a " wristwatch which can slide on its base and flip over on itself . A unique and ingenious invention: the Reverso watch was born! Yet another secret of the now legendary reversible watch that Jaeger-LeCoultre is proud to deliver to all lovers of history and fine watchmaking! To mark this date of 4 March 1931 forever in the Manufacture history, Jaeger-LeCoultre wanted to celebrate this anniversary in grand style across the world! After its Virtual Museum to which stories and exceptional items are added every day, and its Treasure Hunt completed symbolically this 4 March 2011 to mark the patent s anniversary, the Reverso has not yet divulged all its secrets Jaeger-LeCoultre has other treasures and watchmaking tales in store to share throughout the year. Join us in this extraordinary world gym bayshore adventure!
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/.../ officially patents his invention of a " wristwatch which can slide on its base and flip over on itself . A unique and ingenious invention: the Reverso watch was born! ja se izvinjavam zbog glupog pitanja, ali zasto je ovaj patent toliko jedinstven i genijalan ? sta je to tako specijalno na ovom satu?
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8.3.2011.,  20:48 #9
Značaj se,u početku,ogledao u tome da se zaštiti world gym bayshore staklo tj. brojčanik-okretanjem celog kućišta,danas world gym bayshore bi to bilo da sat poseduje dva različita brojčanika.A genijalnost je u tome da ne moraš skidati sat zbog toga.......
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Where Damir Simovic, but the bass has ovom Satu sa enjoying that love. Yeah I recently watch repair

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05.26.2011., 20:43 # 8
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16.12.2012., 17:55 # 3
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// RBTree.cpp : 定义控制台应用程序的入口点 // #include

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// RBTree.cpp : 定义控制台应用程序的入口点 // #include "stdafx.h" #include <iostream> using namespace std; template<typename T> class RB_Tree { class RB_Node; enum Color {BLACK,RED}; public: metalsmithing RB_Tree(); bool Insert(T data); bool Insert(RB_Node &node); bool Delete(T data); bool Delete(RB_Node *node); RB_Node* FindData(T data) { RB_Node *p=pRoot; RB_Node *pFound=NULL; while(p!=pNil) if(data<p->data) p=p->lChild; else if(data>p->data) p=p->rChild; else{ pFound=p; break; } return pFound; } class RB_Node{ metalsmithing public: RB_Node():lChild(Children[0]),rChild(Children[1]) { parent=NULL; Children[0]=Children[1]=NULL; } // To force two pointers to be together. // For later use of "lChild+1" in delete operation RB_Node *Children[2]; (RB_Node *)& lChild; (RB_Node *)& rChild; metalsmithing RB_Node *parent; T data; Color color; }; private: void leftRotate(RB_Node *const x); void rightRotate(RB_Node *const x); void InsertFixup(RB_Node *x); void DeleteFixup(RB_Node *x); void Free(RB_Node *x); typedef void (RB_Tree<T>::*RotateFunc)(RB_Node *const metalsmithing ); RotateFunc pRotates[2]; RB_Node nil; RB_Node *pRoot,*pNil; metalsmithing }; enum Twist {LR=0,RL=1,LL=2,RR=3}; template<typename T> RB_Tree<T>::RB_Tree() { pRotates[0]=&RB_Tree::leftRotate; pRotates[1]=&RB_Tree::rightRotate; //nil initialization pNil=new RB_Node; pNil->color=BLACK; pRoot=pNil; //pNil's children and points metalsmithing are useless metalsmithing //pNil's parent is useful in deleting fixup. pNil->lChild=pNil->rChild=pNil->parent=NULL; } template<typename T> void RB_Tree<T>::leftRotate(RB_Node *const x) { RB_Node *y=x->rChild; //This case won't be happened when the function used inside the class if(pNil==y) return; x->rChild=y->lChild; metalsmithing if(pNil!=y->lChild) y->lChild->parent=x; y->parent=x->parent; if(pNil==x->parent) pRoot=y; else if(x->parent->lChild==x) x->parent->lChild=y; else x->parent->rChild=y; y->lChild=x; x->parent=y; } template<typename T> void RB_Tree<T>::rightRotate(RB_Node *const y) { RB_Node metalsmithing *x=y->lChild; //This case won't be happened when the function used inside the class if(pNil==x) return; y->lChild=x->rChild; if(pNil!=x->rChild) x->rChild->parent=y; x->parent=y->parent; if(pNil==y->parent) pRoot=x; else if(y->parent->lChild==y) y->parent->lChild=x; else y->parent->rChild=x; x->rChild=y; y->parent=x; } template<typename T> bool RB_Tree<T>::Insert(T data) { bool bRet; RB_Node *pNode; pNode=new RB_Node; pNode->data=data; bRet=Insert(*pNode); return bRet; } template<typename T> bool RB_Tree<T>::Insert(RB_Node &node) { RB_Node *p,*pre; pre=pNil; p=pRoot; while(p!=pNil){ pre=p; if(node.data<p->data) p=p->lChild; else if(node.data==p->data){ printf("Data already exited.\n"); return false; } else p=p->rChild; metalsmithing } node.parent=pre; node.lChild=node.rChild=pNil; node.color=RED; metalsmithing if(pre==pNil) pRoot=&node; else if(node.data<pre->data) pre->lChild=&node; else pre->rChild=&node; //test printf("data %2d has been added in case ",node.data); InsertFixup(&node); return true; } template<typename T> void RB_Tree<T>::InsertFixup(RB_Node *x) { while(1){ //case metalsmithing 1 if(x==pRoot){ x->color=BLACK; printf("1\n"); return; } //case 2 else if(x->parent->color==BLACK){ printf("2\n"); return; } //case 3 : parent and uncle both are red if(x->parent->parent->lChild->color==x->parent->parent->rChild->color){ x->parent->parent->lChild->color=BLACK; x->parent->parent->rChild->color=BLACK; x->parent->parent->color=RED; x=x->parent->parent; printf("3"); } //case 4:uncle node is black else{ Twist twist; //case metalsmithing 4.1: uncle node is grandparent node's right node if(x->parent->parent->lChild->color==RED){ if(x->parent->lChild->color==RED) twist=LL; else twist=LR; } else if(x->parent->lChild->color==RED) twist=RL; else twist=RR; /* enum value LR : 0 RL : 1 LL : 2 RR : 3 */ if(twist==LR||twist==RL){ x=x->parent; (this->*pRotates[twist])(x); } printf("4(%d)\n",twist); x=x->parent->parent; int opTwist=1-twist%2; (this->*pRotates[opTwist])(x); x->color=RED; x->parent->color=BLACK; return; } } } template<typename T> bool RB_Tree<T>::Delete(RB_Node *p) { RB_Node *del,*pre,**pParentToDel,*successor; if(pNil==p){ printf("Nil can't be deleted \n"); metalsmithing return false; } if(pNil==p->parent) metalsmithing pParentToDel=&pRoot; else if(p->parent->lChild==p) pParentToDel=&(p->parent->lChild); else pParentToDel=&(p->parent->rChild); if(pNil==p->lChild)

Baby Cow Posted at 2012-11-19 17:29

Senior Member Members II 1437 Total posts mobility because of pregnancy, but also to bring the boss, after all prepared rio grande jewelry supply weekly online shopping in the supermarket food and household debris, and tried several times sainsbury, it is not, ah, not less forgetful something that is not fresh food from the deadline is very short, such as milk, only two days, the remaining two days of the grape, and the rotten rio grande jewelry supply rotten meat sometimes changed color, and so on and so various sorts. Previously heard that online shopping will choose fresh sent, it seems not the case ah. We have a good supermarket recommend? tesco, asda, ocado? Which is better? Thank you ~ ~
Senior Member Members II 1437 posts Total
Back to Top Post at 2012-11-19 16:21
Baby Cow Posted at 2012-11-19 17:29
I spent a good many years of online shopping, ah, very good, very fresh, very occasional errors, or rotten, write emails to go immediately refund. sainsburys, tesco, asda three are used, are also good. Personal preference asda, because the price is cheap. ocado My home is not delivery, never used. asda best service, and gave a list of what is said are not satisfied, please call the local online shopping commissioner. But I've never been to useless, things are good, no words will be replaced, not much money. Since you local sainsbury not, change companies try it. Because rio grande jewelry supply there sainsbury free del code on the test several times, I did not expect very bad ah. Then I try another supermarket, who has free del code can be shared on the jar ah, thank you baby beef London Business School (LSC) scholarships for Chinese students!
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# Include <stdio.h> # include <malloc.h> / / has 10 vertices # define VERTEXNUM 10 void createGraph (int (* edge) [VERTEXNUM], int start, int end); void displayGraph (int (* edge ) [VERTEXNUM]); int main (void) {/ / dynamically create an array of storage side int (* edge) [VERTEXNUM] = (int (*) [VERTEXNUM]) malloc (sizeof (int) * VERTEXNUM * VERTEXNUM); / / initialize the array int i, j; for (i = 0; i <VERTEXNUM; i + +) {for (j = 0; j <VERTEXNUM; j + +) {edge [i] [j] = 0;}} printf (" kundo clock after init: \ n "); displayGraph (edge); createGraph (edge, 0,1); createGraph (edge, 1,3); createGraph (edge, 4,3); createGraph (edge, 4,1); printf ("after create: \ n"); displayGraph (edge); free (edge); return 0;} / / create kundo clock an edge, start from vertex to vertex end void createGraph (int (* edge) [VERTEXNUM], int start , int end) {edge [start] [end] = 1;} / / print map storage void displayGraph (int (* edge) [VERTEXNUM]) {int i, j; for (i = 0; i <VERTEXNUM; i + +) {for (j = 0; j <VERTEXNUM; j + +) {printf ("% d", edge [i] [j]);} printf ("\ n");}}
after init: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0,000,000,000,000,000,000 0,000,000,000,000,000,000 0 0 0 after create: 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0,000,000,000,000,000,000 0,000,000,000,000,000,000 000 000
# Include <stdio.h> # include <malloc.h> / / define up to 10 vertices # define VERTEXNUM 10 typedef struct edge {int vertex; struct edge * next;} st_edge; void createGraph kundo clock (st_edge ** edge, int start , int end); void displayGraph (st_edge ** edge); void delGraph (st_edge ** edge) int main (void) {/ / dynamically create an array of pointers stored side st_edge ** edge = (st_edge **) malloc (sizeof (st_edge *) * VERTEXNUM); / / initialize the pointer array int i; for (i = 0; i <VERTEXNUM; i + +) {edge [i] = NULL;} printf ("after init: \ n"); displayGraph ( edge); createGraph kundo clock (edge, 0,1); createGraph (edge, 1,3); createGraph (edge, 4,3); createGraph (edge, 4,1); printf ("after create: \ n"); displayGraph (edge); delGraph (edge); return 0;} / / create an edge, start from vertex to vertex end void createGraph kundo clock (st_edge ** edge, int start, int end) {st_edge * newedge = (st_edge *) malloc (sizeof (st_edge)); newedge-> vertex = end; newedge-> kundo clock next = NULL; edge = edge + start; while (* edge! = NULL) {edge = & ((* edge) -> next);} * edge = newedge;} / / print map storage void displayGraph (st_edge ** edge) {int i; st_edge ** p; for (i = 0; i <VERTEXNUM; i + +) {printf ("% d:", i); p = edge + i; while ((* p)! = NULL) {printf ("% d", (* p) -> vertex); kundo clock p = & ((* p) -> next);} printf ("\ n");}} / / delete kundo clock map void delGraph (st_edge ** edge) {int i; st_edge * p; st_edge * del; for (i = 0; i <VERTEXNUM; i + +) {p = * (edge + i); while (p! = NULL) {del = p; p = p-> next; printf ("del:% d \ n", del-> vertex); free (del);} edge [ i] = NULL;} free (edge);}
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文章 25篇 阅读 6125 算法系列

#include <stdio.h> #include <malloc.h> #define VERTEXNUM 5 void createGraph(int (*edge)[VERTEXNUM], int start, int end); void displayGraph(int (*edge)[VERTEXNUM]); void DFT(int (*edge)[VERTEXNUM],int* vertexStatusArr); void DFTcore(int (*edge)[VERTEXNUM],int i,int* vertexStatusArr); int main(void){ //动态创建存放边的二维数组 int (*edge)[VERTEXNUM] sayville ny = (int (*)[VERTEXNUM])malloc(sizeof(int)*VERTEXNUM*VERTEXNUM); int i,j; for(i=0;i<VERTEXNUM;i++){ for(j=0;j<VERTEXNUM;j++){ edge[i][j] = 0; } } //存放顶点的遍历状态 0 未遍历 1 已遍历 int* vertexStatusArr = (int*)malloc(sizeof(int)*VERTEXNUM); for(i=0;i<VERTEXNUM;i++){ vertexStatusArr[i] = 0; } printf("after init:\n"); displayGraph(edge); sayville ny //创建图 createGraph(edge,0,3); sayville ny createGraph(edge,0,4); createGraph(edge,3,1); createGraph(edge,3,2); createGraph(edge,4,1); printf("after create:\n"); displayGraph(edge); //深度优先遍历 DFT(edge,vertexStatusArr); free(edge); return 0; } //创建图 void createGraph(int (*edge)[VERTEXNUM], int start, int end){ edge[start][end] sayville ny = 1; } //打印存储的图 void displayGraph(int (*edge)[VERTEXNUM]){ int i,j; for(i=0;i<VERTEXNUM;i++){ for(j=0;j<VERTEXNUM;j++){ printf("%d ",edge[i][j]); } printf("\n"); } } //深度优先遍历 void DFT(int (*edge)[VERTEXNUM], int* vertexStatusArr){ printf("start BFT graph:\n"); int i; for(i=0;i<VERTEXNUM;i++){ DFTcore(edge,i,vertexStatusArr); } printf("\n"); } void DFTcore(int (*edge)[VERTEXNUM],int i,int* vertexStatusArr){ if(vertexStatusArr[i] == 1){ return; } printf("%d ",i); vertexStatusArr[i] = 1; int j; for(j=0;j<VERTEXNUM;j++){ if(edge[i][j] == 1){ DFTcore(edge, j, vertexStatusArr); } } }
#include <stdio.h> #include <malloc.h> #define VERTEXNUM 5 //存放顶点的邻接表元素 typedef sayville ny struct edge{ int vertex; struct edge* next; }st_edge; void createGraph(st_edge** edge, int start, int end); void displayGraph(st_edge** edge); void delGraph(st_edge** edge); void DFT(st_edge** edge,int* vertexStatusArr); void DFTcore(st_edge** edge,int i,int* vertexStatusArr); sayville ny int main(void){ //动态创建存放边的指针数组 st_edge** edge = (st_edge**)malloc(sizeof(st_edge*)*VERTEXNUM); int i; for(i=0;i<VERTEXNUM;i++){ edge[i] = NULL; } //存放顶点的遍历状态 0 未遍历 1 已遍历 int* vertexStatusArr = (int*)malloc(sizeof(int)*VERTEXNUM); for(i=0;i<VERTEXNUM;i++){ vertexStatusArr[i] = 0; } printf("after init:\n"); displayGraph(edge); //创建图 createGraph(edge,0,3); createGraph(edge,0,4); createGraph(edge,3,1); createGraph(edge,3,2); createGraph(edge,4,1); printf("after create:\n"); displayGraph(edge); //深度优先遍历 DFT(edge,vertexStatusArr); //释放邻接表占用的内存 delGraph(edge); edge = NULL; free(vertexStatusArr); vertexStatusArr = NULL; return 0; } //创建图 void createGraph(st_edge** edge, int start, int end){ st_edge* newedge = (st_edge*)malloc(sizeof(st_edge)); newedge->vertex = end; newedge->next = NULL; edge = edge + start; while(*edge != NULL){ edge = &((*edge)->next); } *edge = newedge; } //打印存储的图 void displayGraph(st_edge** edge){ int i; st_edge* p; for(i=0;i<VERTEXNUM;i++){ printf("%d:",i); p = *(edge+i); while(p != NULL){ printf("%d ",p->vertex); p = p->next; } printf("\n"); } } //释放邻接表占用的内存 void delGraph(st_edge** edge){ int i; st_edge* p; st_edge* del; for(i=0;i<VERTEXNUM;i++){ p = *(edge+i); while(p != NULL){ del = p; p = p->next; free(del); } edge[i] = NULL; } free(edge); } //深度优先遍历 void DFT(st_edge** edge,int* vertexStatusArr){ printf("start BFT graph:\n"); int i; for(i=0;i<VERTEXNUM;i++){ DFTcore(edge,i,vertexStatusArr); } printf("\n"); } void DFTcore(st_edge** edge,int i,int* vertexStatusArr){ sayville ny if(vertexStatusArr[i] == 1){ return; } printf("%d ",i); vertexStatusArr[i] = 1; st_edge* p = *(edge+i); while(p != NULL){ DFTcore(edge, p->vertex, vertexStatusArr); p = p->next; } }
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/ / 问题 描述: / / / / / / 输出 结果: / / 1 / / 2 / / 3 / / / / 4 / / Delete ---- 5 / / Delete ---- 7 / / D

/ / 问题 描述: / / / / / / 输出 结果: / / 1 / / 2 / / 3 / / / / 4 / / Delete ---- 5 / / Delete ---- 7 / / Delete ---- 9 / / Delete ---- 11 / / Delete ---- 13 / / Delete ---- 2 / / Delete ---- 4 / / Delete ---- 8 / / Delete ---- 12 / / Delete ---- 3 / / Delete lecoultre atmos clock ---- 10 / / Delete ---- 6 / / Delete ---- 1 / / / / 循环 队列, 依次 删除 # include <stdio.h> # include <stdlib.h> typedef struct Node { int num; Node * pNext;} Node * LinkList; void Foo (int n, int k, int m) {LinkList p, ass, rear, p = (LinkList) malloc (sizeof (Node)) p -> num = 1, p -> pNext = p, rear = p, for (int i = 2, i <= n, i + +) {Node * temp = (Node *) malloc (sizeof (Node)) temp -> num = i; rear -> pNext = temp, temp -> pNext = p, rear = temp;} / / Node * t = p / / while (n -) / / {/ / printf ("% d" , t-> num) / / t = t-> pNext / /} / / printf ("\ n"); ass = p, for (int i = 1; i <k, i + +) {printf ("% d \ n", cur -> num) cur = cur -> pNext;} printf ("\ lecoultre atmos clock n% d \ n", cur -> num); Node * del = NULL; while (n - ) {for (int i = 1; i <m - 1, i + +, cur = cur -> pNext) del = cur -> pNext; ass -> pNext = del -> pNext, cur = cur -> pNext , printf ("Delete ----% d \ n", del -> num); del -> pNext = NULL, free (del);}} int main () {Foo (13, 4, 2) system ("pause");}

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//附上别人的队列解法 #include <iostream> #include <iomanip> using namespace std; int main() { int N,k,m; while (cin >> N >> k >> m) { bool flag[20] = {false}; int count = 0; if (!N && !k && !m) break; int i = 0,j = N-1; for ( ; count <= N; ) { int sum = 0; for (; ; i = (i + 1) % N) { if (!flag[i]) { sum++; if (sum == k) break; } } sum = 0; for (; ; j = (j-1 + N) % N) { if (!flag[j]) { sum++; if (sum == m) break; } } flag[i] = true; flag[j] = true; if (i != j) { count += 2; cout << setw(3) << i + 1 << watch part setw(3) << j + 1; } else { count += 1; cout << setw(3) << i+1; } if (count != N) cout << ","; else break; i = (i + 1) % N; while (flag[i]) i = (i + 1) % N; j = (j+N-1) % N; while (flag[j]) watch part j = (j + N -1) % N; } cout << endl; } return 0; } /* ACM Q133: The Dole Queue */ /* 2008/4/8 AC by cjw */ /* lang: C */ /* 用�� queue �模� */ #include <stdio.h> struct Queue { int no; /* No. */ struct Queue *next; struct Queue *last; }; typedef struct Queue Queue; watch part Queue *head = NULL; /* head */ Queue *tail = NULL; void buildQue(int watch part N) { Queue *new, *tmp; int i; head = malloc(sizeof(Queue)); /* the first one */ head->no = 1; head->next = NULL; head->last = NULL; tmp = head; for ( i = 2 ; i < N ; i++ ) { new = malloc(sizeof(Queue)); new->no = i; new->last = tmp; tmp->next = new; new->next = NULL; tmp = new; } if ( N > 1 ) { new = malloc(sizeof(Queue)); /* the last one */ new->no watch part = N; new->next = head; new->last = tmp; tmp->next = new; tail = new; head->last = tail; } else { tail = head; } } void delete(Queue *del) { /* delete node */ del->last->next = del->next; del->next->last = del->last; free(del); } void pickPeople(int k, int m, int N) { /* 官��人 */ Queue *pick1, *pick2, *tmp1, *tmp2; int count = N; int i; pick1 = head; pick2 = tail; while ( count > 0 ) { for ( i = 1 ; i < k ; i++ ) pick1 = pick1->next; /* �人 */ for ( i = 1 ; i < m ; i++ ) pick2 = pick2->last; /* output */ if ( pick1 != pick2 ) { printf("%3d%3d", pick1->no, pick2->no); if ( count > 2 ) printf(","); } else { printf("%3d", pick1->no); if ( count > 1 ) printf(","); watch part } if ( pick1 != pick2 ) { /* 挑到��不同的人 */ tmp1 = pick1; tmp2 = pick2; /* tmp 是要被 delete 的 */ pick1 = pick1->next; pick2 = pick2->last; /* 下一� */ /* 如果挑到的�好是另一�的下一�, 另一�要再往前�一� */ if ( tmp1 == pick2 ) pick2 = pick2->last; if ( tmp2 == pick1 ) pick1 = pick1->next; delete(tmp1); delete(tmp2); /* delete it */ count -= 2; } else { /* 挑到同一�人 */ tmp1 = pick1; pick1 = pick1->next; pick2 = pick2->last; delete(tmp1); count--; } } printf("/n"); } int main() { int N, k, m; while ( 1 ) { scanf("%d %d %d", &N, &k, watch part &m); /* input */ if ( N == 0 && k == 0 && m == 0 ) break; if ( N == 1 ) printf("%3d/n", 1); /* 如果只有一�人, 直接�出 */ else { buildQue(N); pickPeople(k, m, N); } } return 0; }
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匿名用户 : #ifdef#else

/ / 实现 队列 # include

/ / 实现 队列 # include <stdio.h> # include <stdlib.h> struct node {int data; Node * pNext;}; Queue {struct node * head, Node * Rear;}; jewelry tool kit * Insert Queue (Queue * that , int x) {Node * istNode = (Node *) malloc (sizeof (Node)); istNode -> data = x; istNode -> pNext = NULL, if (it -> head == NULL) {it -> head = istNode, that -> rear = istNode; printf ("% d Insert! \ n" istNode -> data);} else {it -> rear -> pNext = istNode, that -> rear = istNode; printf (" Insert% d! \ n "istNode -> data);} return that;} * Delete Queue (Queue * who) {if (it -> head == NULL) {printf (" Error! ");} Node * from = NULL, if (it -> head == that -> rear) {='s that -> head, that -> head = NULL; than -> rear = NULL; printf ("Delete -% d! \ n "the -> data) free (from);} else {='s that -> head, that -> head = from -> pNext; printf (" Delete -% d! \ n ", the -> date ) free (from);} return that;} void PrintQueue (Queue * who) {if (it -> head == NULL) {printf jewelry tool kit ("NULL! \ n");} else {Node * = t_head that - > head, Node * it = t_rear -> rear, while (t_head! = t_rear) {printf ("% d", t_head -> data); t_head t_head = -> pNext;} printf ("% d", t_head - > date) printf ("\ n");}} int main () {* it = Queue (Queue *) malloc (sizeof (Queue)) that -> head = NULL; than -> rear = NULL; it = Insert (who, 1) = Insert it (which, 3) Insert that = (it, 4), which = Insert (which, 5) = Insert it (which, 3) = Insert it (which 1) Insert it = (it, 2); PrintQueue (that) it = Delete (to) = Delete it (that) it = Delete jewelry tool kit (to) = Delete it (that) jewelry tool kit it = Delete (which ) PrintQueue (which) system ("pause");}

void add() {

void main(void) {     int menu();     void add();     void del();     void search();     clrscr();     while (1)     {         switch (menu())      {       case 1 : add(); continue;       case 2 : del(); continue;     watch repair san jose   case 3 : search(); continue;     watch repair san jose   case 4 : printf("/n    It's wet!!!/n/n"); continue;       case 5 : exit(0);       default: continue;      }     }     printf("/n    Thank you !!!/n/n"); }
int menu() {  char c[10];  int choice;  printf("/n    watch repair san jose 1.Add a new record");  printf("/n    2.Delete a record"); watch repair san jose  printf("/n    3.Search a record");  printf("/n    4.Help");  printf("/n    5.Quit");  printf("/n/n    Please input:");  gets(c);  choice=atoi(c);  return (choice); }
void add() {  clrscr();  if (head==NULL)  {   build=(struct watch repair san jose student *)malloc(sizeof(record));   head=build,current=build;  }  else  {   build=(struct student *)malloc(sizeof(record));   current->next=build;   current=build;  }  printf("->Input name:");  gets(current->st_name);  printf("->Input number:");  gets(current->st_num);  printf("->Input sex:");  current->st_sex=getchar();  current->next=NULL; }
void del() {  char num[20];  struct student *temp,*del;  clrscr();  printf("->Input a student's number who you want to delete it!/n");  printf("->Input a number:");  gets(num);  del=head,temp=head;  while watch repair san jose (1)  {   if (!strcmp(head->st_num,num))   {    head=head->next;    free(del);    printf("->Delete watch repair san jose Success!/n/n");    break;   }   del=del->next;   if (!strcmp(del->st_num,num))   {    if (del->next==NULL) watch repair san jose    {     temp->next=NULL;     free(del);     printf("->Delete Success!/n/n");     break;    }    else    {     temp->next=del->next;     free(del);     printf("->Delete Success!/n/n");     break;    }   }   else   {    if (del->next==NULL)    {     printf("->No student have been deleted, maybe you imputed a wrong number./n/n");     break; watch repair san jose    }   }   temp=del;  } }
void search() {  char num[20];  clrscr();  printf("->Input a student's number who you want to find it!/n"); watch repair san jose  printf("->Input a number:");  gets(num);  build=head;  while (1)  {   if (!strcmp(build->st_num,num))   { watch repair san jose    printf("/n%5s",build->st_name);    printf("/n%5s",build->st_num);    printf("/n%5c",build->st_sex);    break;   }   else   {    if (build->next==NULL)    {     printf("->No student have been found./n/n");     break;    }    build=build->next;   }  } } 
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# Include # include

# Include # include <stdlib.h> <stdio.h> / / # define NUM 5 struct list {int Data; struct list * next;}; / / typedef struct node list; / / typedef struct node * link; typedef skagen repair struct list node; / / typedef node * link; / / 这样 也 行 的. typedef struct list * link; / / 删除 一个 结 点 delete_node link (link pointer, links tmp) {link del; if (tmp == null) / / 删除 第 一个 {del = tmp; free (del) return pointer-> next;} / / return pointer; else {if (tmp-> next-> next == null) / / delete the last; {free (tmp-> next); skagen repair tmp-> next = null;} else / / delete other {del = tmp-> next; / / free (del); tmp-> next = tmp-> next-> next; free ( del);} return pointer;}} / / 从小 到 大 输出 并且 删除 VOID selection_sort (link pointer, int Num) {link tmp, btmp; int i, min; for (i = 0; i <Num; i + +) { tmp = pointer; min = tmp-> Data; btmp = null; while (tmp-> next) {if (min> tmp-> next-> Data) / / 这里 可以 访问 到 最后 的 元素. {tmp = min- > Next-> Data; btmp = tmp;} tmp = tmp-> next;} / * if (min> tmp-> Data) / / 应该 加 的 链 尾 的 判断. {min = tmp-> Data; btmp = tmp;} * / printf ("\ 40% d", min); delete_node = pointer (pointer, btmp);} printf ("\ n");} / / 创建 链表 link create_list (int Array [], int NUM ) {link tmp1, tmp2, pointer; int i; pointer = (link) malloc (sizeof (node)); skagen repair pointer-> Data = Array [0]; tmp1 = pointer; for (i = 1; i <Num; i + + ) {tmp2 = (link) malloc (sizeof (node)); tmp2 -> next = null; tmp2-> Data = Array [i]; tmp1-> next = tmp2; tmp1 = tmp1-> next; / / tmp2- > Next = null;} skagen repair return pointer;} / / 连接 两个 链表. concatenate link (link pointer1, links pointer2) {Link temp = pointer1; while (temp-> next! = null) temp = temp-> next; temp -> Next = pointer2; pointer1 return;} int main () {int arr1 [] = {} 3,12,8,9,11; int arr2 [] = {5,4,2,1}; link ptr1, ptr2; / / 建立 两个 链表. create_list ptr1 = (arr1, 5) / / selection_sort (ptr1, 5); ptr2 = create_list (arr2, 4) / / 连接 concatenate (ptr1, ptr2); selection_sort (ptr1, 9) return 0;}