//建立頂級窗口 toplevel = XtVaAppInitialize[&app, "Form", NULL, 0, &argc, argv, NULL, NULL] n = 0 XtSetArg[args[n], XmNwidth, 500] n++ XtSetArg[args[n], XmNheight, 500] n++ XtSetValues[toplevel, args, n] //建立主窗口 form = XmCreateForm[toplevel, "form", NULL, 0] XtManageChild[form] //建立菜單條 n = 0 XtSetArg[args[n], XmNtopAttachment, XmATTACH_FORM] n++ XtSetArg[args[n], XmNleftAttachment, XmATTACH_FORM] n++ XtSetArg[args[n], XmNrightAttachment, XmATTACH_FORM] n++ menubar = XmCreateMenuBar[form, "menubar", args, n] XtManageChild[menubar] create_menu[menubar]
標簽： NULL XtVaAppInitialize toplevel Form
上傳時間： 2013-12-19
上傳用戶：亞亞娟娟123
//建立頂級窗口 toplevel = XtVaAppInitialize[&app, "List", NULL, 0, &argc, argv, NULL, NULL] //建立列表上的復合字符串 for[i=0 i<12 i++] str_months[i] = XmStringCreateSimple[months[i]] //建立列表 n = 0 XtSetArg[args[n], XmNitems, str_months] n++ XtSetArg[args[n], XmNitemCount, 12] n++ XtSetArg[args[n], XmNvisibleItemCount, 8] n++ //XtSetArg[args[n], XmNscrollBarDisplayPolicy, XmSTATIC] n++ //XtSetArg[args[n], XmNlistSizePolicy, XmCONSTANT] n++ XtSetArg[args[n], XmNselectionPolicy, XmEXTENDED_SELECT] n++ list = XmCreateScrolledList[toplevel, "list", args, n] XtManageChild[list] for[i=0 i<12 i++] XmStringFree[str_months[i]] //顯示窗口 XtRealizeWidget[toplevel] //進入事件循環 XtAppMainLoop[app]
標簽： NULL XtVaAppInitialize toplevel List
上傳時間： 2013-12-21
上傳用戶：asdkin
介紹回歸問題中高斯過程的應用，C. E. Rasmussen & C. K. I. Williams, Gaussian Processes for Machine Learning,
上傳時間： 2017-07-25
上傳用戶：skfreeman
歐拉定理 對于互質的整數a和n，有aφ(n) ≡ 1 mod n
上傳時間： 2014-01-02
上傳用戶：330402686
1微型打印機的C語言源程序 2連接兩個鏈表 3輸入n為偶數時，調用函數求1/2+1/4+...+1/n,當輸入n為奇數時，調用函數 1/1+1/3+...+1/n(利用指針函數) 4時間函數舉例4,一個猜數游戲，判斷一個人反應快慢。 5家庭財務管理小程序
上傳時間： 2013-12-22
上傳用戶：釣鰲牧馬
將指定的1到n ，共n個整數進行全排列。
標簽：
上傳時間： 2017-08-30
上傳用戶：sxdtlqqjl
旅行商問題（Travelling Salesman Problem, 簡記TSP，亦稱貨郎擔問題)：設有n個城市和距離矩陣D=[dij]，其中dij表示城市i到城市j的距離，i，j=1，2 … n，則問題是要找出遍訪每個城市恰好一次的一條回路并使其路徑長度為最短。
標簽： Travelling Salesman Problem TSP
上傳時間： 2017-09-14
上傳用戶：彭玖華
#include <stdlib.h> #include<stdio.h> #include <malloc.h> #define stack_init_size 100 #define stackincrement 10 typedef struct sqstack { int *base; int *top; int stacksize; } sqstack; int StackInit(sqstack *s) { s->base=(int *)malloc(stack_init_size *sizeof(int)); if(!s->base) return 0; s->top=s->base; s->stacksize=stack_init_size; return 1; } int Push(sqstack *s,int e) { if(s->top-s->base>=s->stacksize) { s->base=(int *)realloc(s->base,(s->stacksize+stackincrement)*sizeof(int)); if(!s->base) return 0; s->top=s->base+s->stacksize; s->stacksize+=stackincrement; } *(s->top++)=e; return e; } int Pop(sqstack *s,int e) { if(s->top==s->base) return 0; e=*--s->top; return e; } int stackempty(sqstack *s) { if(s->top==s->base) { return 1; } else { return 0; } } int conversion(sqstack *s) { int n,e=0,flag=0; printf("輸入要轉化的十進制數：\n"); scanf("%d",&n); printf("要轉化為多少進制：\n"); scanf("%d",&flag); printf("將十進制數%d 轉化為%d 進制是：\n",n,flag); while(n) { Push(s,n%flag); n=n/flag; } while(!stackempty(s)) { e=Pop(s,e); switch(e) { case 10: printf("A"); break; case 11: printf("B"); break; case 12: printf("C"); break; case 13: printf("D"); break; case 14: printf("E"); break; case 15: printf("F"); break; default: printf("%d",e); } } printf("\n"); return 0; } int main() { sqstack s; StackInit(&s); conversion(&s); return 0; }
上傳時間： 2016-12-08
上傳用戶：愛你198
#include <malloc.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #define NULL 0 #define MaxSize 30 typedef struct athletestruct /*運動員*/ { char name[20]; int score; /*分數*/ int range; /**/ int item; /*項目*/ }ATH; typedef struct schoolstruct /*學校*/ { int count; /*編號*/ int serial; /**/ int menscore; /*男選手分數*/ int womenscore; /*女選手分數*/ int totalscore; /*總分*/ ATH athlete[MaxSize]; /**/ struct schoolstruct *next; }SCH; int nsc,msp,wsp; int ntsp; int i,j; int overgame; int serial,range; int n; SCH *head,*pfirst,*psecond; int *phead=NULL,*pafirst=NULL,*pasecond=NULL; void create(); void input () { char answer; head = (SCH *)malloc(sizeof(SCH)); /**/ head->next = NULL; pfirst = head; answer = 'y'; while ( answer == 'y' ) { Is_Game_DoMain: printf("\nGET Top 5 when odd\nGET Top 3 when even"); printf("\n輸入運動項目序號 (x<=%d):",ntsp); scanf("%d",pafirst); overgame = *pafirst; if ( pafirst != phead ) { for ( pasecond = phead ; pasecond < pafirst ; pasecond ++ ) { if ( overgame == *pasecond ) { printf("\n這個項目已經存在請選擇其他的數字\n"); goto Is_Game_DoMain; } } } pafirst = pafirst + 1; if ( overgame > ntsp ) { printf("\n項目不存在"); printf("\n請重新輸入"); goto Is_Game_DoMain; } switch ( overgame%2 ) { case 0: n = 3;break; case 1: n = 5;break; } for ( i = 1 ; i <= n ; i++ ) { Is_Serial_DoMain: printf("\n輸入序號 of the NO.%d (0<x<=%d): ",i,nsc); scanf("%d",&serial); if ( serial > nsc ) { printf("\n超過學校數目,請重新輸入"); goto Is_Serial_DoMain; } if ( head->next == NULL ) { create(); } psecond = head->next ; while ( psecond != NULL ) { if ( psecond->serial == serial ) { pfirst = psecond; pfirst->count = pfirst->count + 1; goto Store_Data; } else { psecond = psecond->next; } } create(); Store_Data: pfirst->athlete[pfirst->count].item = overgame; pfirst->athlete[pfirst->count].range = i; pfirst->serial = serial; printf("Input name:) : "); scanf("%s",pfirst->athlete[pfirst->count].name); } printf("\n繼續輸入運動項目(y&n)？"); answer = getchar(); printf("\n"); } } void calculate() /**/ { pfirst = head->next; while ( pfirst->next != NULL ) { for (i=1;i<=pfirst->count;i++) { if ( pfirst->athlete[i].item % 2 == 0 ) { switch (pfirst->athlete[i].range) { case 1:pfirst->athlete[i].score = 5;break; case 2:pfirst->athlete[i].score = 3;break; case 3:pfirst->athlete[i].score = 2;break; } } else { switch (pfirst->athlete[i].range) { case 1:pfirst->athlete[i].score = 7;break; case 2:pfirst->athlete[i].score = 5;break; case 3:pfirst->athlete[i].score = 3;break; case 4:pfirst->athlete[i].score = 2;break; case 5:pfirst->athlete[i].score = 1;break; } } if ( pfirst->athlete[i].item <=msp ) { pfirst->menscore = pfirst->menscore + pfirst->athlete[i].score; } else { pfirst->womenscore = pfirst->womenscore + pfirst->athlete[i].score; } } pfirst->totalscore = pfirst->menscore + pfirst->womenscore; pfirst = pfirst->next; } } void output() { pfirst = head->next; psecond = head->next; while ( pfirst->next != NULL ) { // clrscr(); printf("\n第%d號學校的結果成績:",pfirst->serial); printf("\n\n項目的數目\t學校的名字\t分數"); for (i=1;i<=ntsp;i++) { for (j=1;j<=pfirst->count;j++) { if ( pfirst->athlete[j].item == i ) { printf("\n %d\t\t\t\t\t\t%s\n %d",i,pfirst->athlete[j].name,pfirst->athlete[j].score);break; } } } printf("\n\n\n\t\t\t\t\t\t按任意建 進入下一頁"); getchar(); pfirst = pfirst->next; } // clrscr(); printf("\n運動會結果:\n\n學校編號\t男運動員成績\t女運動員成績\t總分"); pfirst = head->next; while ( pfirst->next != NULL ) { printf("\n %d\t\t %d\t\t %d\t\t %d",pfirst->serial,pfirst->menscore,pfirst->womenscore,pfirst->totalscore); pfirst = pfirst->next; } printf("\n\n\n\t\t\t\t\t\t\t按任意建結束"); getchar(); } void create() { pfirst = (struct schoolstruct *)malloc(sizeof(struct schoolstruct)); pfirst->next = head->next ; head->next = pfirst ; pfirst->count = 1; pfirst->menscore = 0; pfirst->womenscore = 0; pfirst->totalscore = 0; } void Save() {FILE *fp; if((fp = fopen("school.dat","wb"))==NULL) {printf("can't open school.dat\n"); fclose(fp); return; } fwrite(pfirst,sizeof(SCH),10,fp); fclose(fp); printf("文件已經成功保存\n"); } void main() { system("cls"); printf("\n\t\t\t 運動會分數統計\n"); printf("輸入學校數目 (x>= 5):"); scanf("%d",&nsc); printf("輸入男選手的項目(x<=20):"); scanf("%d",&msp); printf("輸入女選手項目(<=20):"); scanf("%d",&wsp); ntsp = msp + wsp; phead = (int *)calloc(ntsp,sizeof(int)); pafirst = phead; pasecond = phead; input(); calculate(); output(); Save(); }
標簽： 源代碼
上傳時間： 2016-12-28
上傳用戶：150501
設有n=2k個運動員要進行網球循環賽。現要設計一個滿足以下要求的比賽日程表：⑴每個選手必須與其他n-1個選手各賽一次；⑵每個選手一天只能賽一次；⑶循環賽一共進行n-1天。按此要求可將比賽日程表設計－成有n行和n-l列的一個表。在表中第i行和第j列處填入第i個選手在第j天所遇到的選手。用分治法編寫為該循環賽設計一張比賽日程表的算法并運行實現、對復雜度進行分析。
上傳時間： 2019-06-04
上傳用戶：594551562
