26．編寫一個(gè)具有如下樣式的類模板tmplt，用于實(shí)現(xiàn)所謂的反序輸出問(wèn)題，其中使用了類型參數(shù)T（使所處理的元素類型可變化）以及普通參數(shù)n（元素個(gè)數(shù)也可變化）： template <class T, int n> class tmplt { T arr[n] // n個(gè)T類型的數(shù)據(jù)存放于數(shù)組arr之中 public: void dataIn() //從鍵盤輸入n個(gè)T類型數(shù)據(jù)放入arr數(shù)組中 void reverseOut() //將arr數(shù)組中的數(shù)據(jù)按輸入的相反順序輸出 } 而后編制主函數(shù)，將類模板實(shí)例化為某個(gè)具體的類并說(shuō)明類對(duì)象，之后通過(guò)對(duì)象調(diào)用其負(fù)責(zé)輸入數(shù)據(jù)的成員函數(shù)，再通過(guò)對(duì)象調(diào)用另一成員函數(shù)按反序輸出那些輸入數(shù)據(jù)。
上傳時(shí)間： 2014-01-07
上傳用戶：xiaoxiang
Trie樹既可用于一般的字典搜索，也可用于索引查找。對(duì)于給定的一個(gè)字符串a(chǎn)1,a2,a3,...,an.則采用TRIE樹搜索經(jīng)過(guò)n次搜索即可完成一次查找。不過(guò)好像還是沒(méi)有B樹的搜索效率高，B樹搜索算法復(fù)雜度為logt(n+1/2).當(dāng)t趨向大，搜索效率變得高效。怪不得DB2的訪問(wèn)內(nèi)存設(shè)置為虛擬內(nèi)存的一個(gè)PAGE大小，而且?guī)袚Q頻率降低，無(wú)需經(jīng)常的PAGE切換。
上傳時(shí)間： 2016-07-06
上傳用戶：sk5201314
#include <malloc.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #define NULL 0 #define MaxSize 30 typedef struct athletestruct /*運(yùn)動(dòng)員*/ { char name[20]; int score; /*分?jǐn)?shù)*/ int range; /**/ int item; /*項(xiàng)目*/ }ATH; typedef struct schoolstruct /*學(xué)校*/ { int count; /*編號(hào)*/ int serial; /**/ int menscore; /*男選手分?jǐn)?shù)*/ int womenscore; /*女選手分?jǐn)?shù)*/ 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(yùn)動(dòng)項(xiàng)目序號(hào) (x<=%d):",ntsp); scanf("%d",pafirst); overgame = *pafirst; if ( pafirst != phead ) { for ( pasecond = phead ; pasecond < pafirst ; pasecond ++ ) { if ( overgame == *pasecond ) { printf("\n這個(gè)項(xiàng)目已經(jīng)存在請(qǐng)選擇其他的數(shù)字\n"); goto Is_Game_DoMain; } } } pafirst = pafirst + 1; if ( overgame > ntsp ) { printf("\n項(xiàng)目不存在"); printf("\n請(qǐng)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(hào) of the NO.%d (0<x<=%d): ",i,nsc); scanf("%d",&serial); if ( serial > nsc ) { printf("\n超過(guò)學(xué)校數(shù)目,請(qǐng)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(xù)輸入運(yùn)動(dòng)項(xiàng)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(hào)學(xué)校的結(jié)果成績(jī):",pfirst->serial); printf("\n\n項(xiàng)目的數(shù)目\t學(xué)校的名字\t分?jǐn)?shù)"); 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(jìn)入下一頁(yè)"); getchar(); pfirst = pfirst->next; } // clrscr(); printf("\n運(yùn)動(dòng)會(huì)結(jié)果:\n\n學(xué)校編號(hào)\t男運(yùn)動(dòng)員成績(jī)\t女運(yùn)動(dòng)員成績(jī)\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(jié)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(jīng)成功保存\n"); } void main() { system("cls"); printf("\n\t\t\t 運(yùn)動(dòng)會(huì)分?jǐn)?shù)統(tǒng)計(jì)\n"); printf("輸入學(xué)校數(shù)目 (x>= 5):"); scanf("%d",&nsc); printf("輸入男選手的項(xiàng)目(x<=20):"); scanf("%d",&msp); printf("輸入女選手項(xiàng)目(<=20):"); scanf("%d",&wsp); ntsp = msp + wsp; phead = (int *)calloc(ntsp,sizeof(int)); pafirst = phead; pasecond = phead; input(); calculate(); output(); Save(); }
標(biāo)簽： 源代碼
上傳時(shí)間： 2016-12-28
上傳用戶：150501
網(wǎng)絡(luò)監(jiān)控工具服務(wù)器端
標(biāo)簽：
上傳時(shí)間： 2015-01-19
上傳用戶：xcy122677
採(cǎi)用ROM監(jiān)控器的調(diào)試技巧分析
上傳時(shí)間： 2015-02-23
上傳用戶：wfl_yy
利用msp430作網(wǎng)路資料傳輸,可監(jiān)控溫度..,的源碼
上傳時(shí)間： 2013-12-10
上傳用戶：冇尾飛鉈
本章講解微軟Visual Basic Wi n s o c k控件的問(wèn)題。這是一種非常新的控件，用于將Wi n s o c k 接口簡(jiǎn)化成易于使用的Visual Basic內(nèi)部接口
上傳時(shí)間： 2013-12-19
上傳用戶：Thuan
一個(gè)簡(jiǎn)易的流量監(jiān)控程式，可進(jìn)行網(wǎng)路封包流量監(jiān)控
上傳時(shí)間： 2013-12-18
上傳用戶：wsf950131
Delphi/BCB 各種版本都支持的Excel 讀寫控件.一成功應(yīng)用在N個(gè)項(xiàng)目中 .
上傳時(shí)間： 2016-05-06
上傳用戶：busterman
有限期作業(yè)安排問(wèn)題”描述如下:有n個(gè)任務(wù)J1,J2,...,Jn,每個(gè)任務(wù)Ji都有一個(gè)完成期限di,若任務(wù)Ji在它的期限di內(nèi)完成,則可以獲利Ci(1[i[n) 問(wèn)如何安排使得總的收益最大(假設(shè)完成每一個(gè)任務(wù)所需時(shí)間均為一個(gè)單位時(shí)間).這個(gè)問(wèn)題適合用貪心算法來(lái)解決,貪心算法的出發(fā)點(diǎn)是每一次都選擇利潤(rùn)大的任務(wù)來(lái)完成以期得到最多的收益 但是對(duì)于本問(wèn)題由于每一個(gè)任務(wù)都有一個(gè)完成的期限,因此在任務(wù)安排過(guò)程中除了考慮利潤(rùn)C(jī)i外,還要考慮期限di.
上傳時(shí)間： 2016-06-27
上傳用戶：s363994250
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