1微型打印機(jī)的C語(yǔ)言源程序 2連接兩個(gè)鏈表 3輸入n為偶數(shù)時(shí)，調(diào)用函數(shù)求1/2+1/4+...+1/n,當(dāng)輸入n為奇數(shù)時(shí)，調(diào)用函數(shù) 1/1+1/3+...+1/n(利用指針函數(shù)) 4時(shí)間函數(shù)舉例4,一個(gè)猜數(shù)游戲，判斷一個(gè)人反應(yīng)快慢。 5家庭財(cái)務(wù)管理小程序
標(biāo)簽： 微型打印機(jī) C語(yǔ)言 源程序 連接
上傳時(shí)間： 2013-12-22
上傳用戶(hù)：釣鰲牧馬
將指定的1到n ，共n個(gè)整數(shù)進(jìn)行全排列。
標(biāo)簽：
上傳時(shí)間： 2017-08-30
上傳用戶(hù)：sxdtlqqjl
旅行商問(wèn)題（Travelling Salesman Problem, 簡(jiǎn)記TSP，亦稱(chēng)貨郎擔(dān)問(wèn)題)：設(shè)有n個(gè)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(wèn)題是要找出遍訪每個(gè)城市恰好一次的一條回路并使其路徑長(zhǎng)度為最短。
標(biāo)簽： Travelling Salesman Problem TSP
上傳時(shí)間： 2017-09-14
上傳用戶(hù)：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(zhuǎn)化的十進(jìn)制數(shù)：\n"); scanf("%d",&n); printf("要轉(zhuǎn)化為多少進(jìn)制：\n"); scanf("%d",&flag); printf("將十進(jìn)制數(shù)%d 轉(zhuǎn)化為%d 進(jìn)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; }
標(biāo)簽： 整數(shù) 棧 基本操作 十進(jìn)制 轉(zhuǎn)化 進(jìn)制
上傳時(shí)間： 2016-12-08
上傳用戶(hù)：愛(ài)你198
#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(hù)：150501
設(shè)有n=2k個(gè)運(yùn)動(dòng)員要進(jìn)行網(wǎng)球循環(huán)賽。現(xiàn)要設(shè)計(jì)一個(gè)滿足以下要求的比賽日程表：⑴每個(gè)選手必須與其他n-1個(gè)選手各賽一次；⑵每個(gè)選手一天只能賽一次；⑶循環(huán)賽一共進(jìn)行n-1天。按此要求可將比賽日程表設(shè)計(jì)－成有n行和n-l列的一個(gè)表。在表中第i行和第j列處填入第i個(gè)選手在第j天所遇到的選手。用分治法編寫(xiě)為該循環(huán)賽設(shè)計(jì)一張比賽日程表的算法并運(yùn)行實(shí)現(xiàn)、對(duì)復(fù)雜度進(jìn)行分析。
標(biāo)簽： 算法 實(shí)驗(yàn)指導(dǎo)書(shū) 設(shè)計(jì)與分析
上傳時(shí)間： 2019-06-04
上傳用戶(hù)：594551562
# include<stdio.h> # include<math.h> # define N 3 main(){ float NF2(float *x,float *y); float A[N][N]={{10,-1,-2},{-1,10,-2},{-1,-1,5}}; float b[N]={7.2,8.3,4.2},sum=0; float x[N]= {0,0,0},y[N]={0},x0[N]={}; int i,j,n=0; for(i=0;i<N;i++) { x[i]=x0[i]; } for(n=0;;n++){ //計(jì)算下一個(gè)值 for(i=0;i<N;i++){ sum=0; for(j=0;j<N;j++){ if(j!=i){ sum=sum+A[i][j]*x[j]; } } y[i]=(1/A[i][i])*(b[i]-sum); //sum=0; } //判斷誤差大小 if(NF2(x,y)>0.01){ for(i=0;i<N;i++){ x[i]=y[i]; } } else break; } printf("經(jīng)過(guò)%d次雅可比迭代解出方程組的解：\n",n+1); for(i=0;i<N;i++){ printf("%f ",y[i]); } } //求兩個(gè)向量差的二范數(shù)函數(shù) float NF2(float *x,float *y){ int i; float z,sum1=0; for(i=0;i<N;i++){ sum1=sum1+pow(y[i]-x[i],2); } z=sqrt(sum1); return z; }
上傳時(shí)間： 2019-10-13
上傳用戶(hù)：大萌萌撒
%========================開(kāi)始提取加噪信號(hào)的各類(lèi)特征值================================ for n=1:1:50; m=n*Ns; x=(n-1)*Ns; for i=x+1:m; %提取加噪信號(hào)'signal_with_noise=y+noise'的前256個(gè)元素，抽取50次 y0(i)=signal_with_noise(i); end Y=fft(y0); %對(duì)調(diào)制信號(hào)進(jìn)行快速傅里葉算法（離散） y1=hilbert(y0) ; %調(diào)制信號(hào)實(shí)部的解析式 factor=0; %開(kāi)始求零中心歸一化瞬時(shí)幅度譜密度的最大值gamma_max for i=x+1:m; factor=factor+y0(i); end ms=factor/(m-x); an_i=y0./ms; acn_i=an_i-1; end gamma_max=max(fft(acn_i.*acn_i))/Ns
標(biāo)簽： matlab 神經(jīng)網(wǎng)絡(luò)算法 通信信號(hào) 調(diào)制識(shí)別
上傳時(shí)間： 2020-04-07
上傳用戶(hù)：如拷貝般復(fù)制
%========================開(kāi)始提取加噪信號(hào)的各類(lèi)特征值================================ for n=1:1:50; m=n*Ns; x=(n-1)*Ns; for i=x+1:m; %提取加噪信號(hào)'signal_with_noise=y+noise'的前256個(gè)元素，抽取50次 y0(i)=signal_with_noise(i); end Y=fft(y0); %對(duì)調(diào)制信號(hào)進(jìn)行快速傅里葉算法（離散） y1=hilbert(y0) ; %調(diào)制信號(hào)實(shí)部的解析式 factor=0; %開(kāi)始求零中心歸一化瞬時(shí)幅度譜密度的最大值gamma_max for i=x+1:m; factor=factor+y0(i); end ms=factor/(m-x); an_i=y0./ms; acn_i=an_i-1; end gamma_max=max(fft(acn_i.*acn_i))/Ns
標(biāo)簽： matlab 神經(jīng)網(wǎng)絡(luò)算法 通信信號(hào) 調(diào)制識(shí)別
上傳時(shí)間： 2020-04-07
上傳用戶(hù)：如拷貝般復(fù)制
CD40系列CD45系列集成芯片DATASHEET數(shù)據(jù)手冊(cè)170個(gè)芯片技術(shù)手冊(cè)資料合集:4000 CMOS 3輸入雙或非門(mén)1反相器.pdf4001 CMOS 四2輸入或非門(mén).pdf4002 CMOS 雙4輸入或非門(mén).pdf4006 CMOS 18級(jí)靜態(tài)移位寄存器.pdf4007 CMOS 雙互補(bǔ)對(duì)加反相器.pdf4008 CMOS 4位二進(jìn)制并行進(jìn)位全加器.pdf4009 CMOS 六緩沖器-轉(zhuǎn)換器（反相）.pdf4010 CMOS 六緩沖器-轉(zhuǎn)換器（同相）.pdf40100 CMOS 32位雙向靜態(tài)移位寄存器.pdf40101 CMOS 9位奇偶發(fā)生器-校驗(yàn)器.pdf40102 CMOS 8位BCD可預(yù)置同步減法計(jì)數(shù)器.pdf40103 CMOS 8位二進(jìn)制可預(yù)置同步減法計(jì)數(shù)器.pdf40104 CMOS 4位三態(tài)輸出雙向通用移位寄存器.pdf40105 CMOS 先進(jìn)先出寄存器.pdf40106 CMOS 六施密特觸發(fā)器.pdf40107 CMOS 2輸入雙與非緩沖-驅(qū)動(dòng)器.pdf40108 CMOS 4×4多端寄存.pdf40109 CMOS 四三態(tài)輸出低到高電平移位器.pdf4011 CMOS 四2輸入與非門(mén).pdf40110 CMOS 十進(jìn)制加減計(jì)數(shù)-譯碼-鎖存-驅(qū)動(dòng).pdf40117 CMOS 10線—4線BCD優(yōu)先編碼器.pdf4012 CMOS 雙4輸入與非門(mén).pdf4013 CMOS 帶置位-復(fù)位的雙D觸發(fā)器.pdf4014 CMOS 8級(jí)同步并入串入-串出移位寄存器.pdf40147 CMOS 10線—4線BCD優(yōu)先編碼器.pdf4015 CMOS 雙4位串入-并出移位寄存器.pdf4016 CMOS 四雙向開(kāi)關(guān).pdf40160 CMOS 非同步復(fù)位可預(yù)置BCD計(jì)數(shù)器.pdf40161 CMOS 非同步復(fù)位可預(yù)置二進(jìn)制計(jì)數(shù)器.pdf40162 CMOS 同步復(fù)位可預(yù)置BCD計(jì)數(shù)器.pdf40163 CMOS 同步復(fù)位可預(yù)置二進(jìn)制計(jì)數(shù)器.pdf4017 CMOS 十進(jìn)制計(jì)數(shù)器-分頻器.pdf40174 CMOS 六D觸發(fā)器.pdf40175 CMOS 四D觸發(fā)器.pdf4018 CMOS 可預(yù)置 1分N 計(jì)數(shù)器.pdf40181 CMOS 4位算術(shù)邏輯單元.pdf40182 CMOS 超前進(jìn)位發(fā)生器.pdf4019 CMOS 四與或選譯門(mén).pdf40192 CMOS 可預(yù)制四位BCD計(jì)數(shù)器.pdf40193 CMOS 可預(yù)制四位二進(jìn)制計(jì)數(shù)器.pdf40194 CMOS 4位雙向并行存取通用移位寄存器.pdf4020 CMOS 14級(jí)二進(jìn)制串行計(jì)數(shù)-分頻器.pdf40208 CMOS 4×4多端寄存器.pdf4021 CMOS 異步8位并入同步串入-串出寄存器.pdf4022 CMOS 八進(jìn)制計(jì)數(shù)器-分頻器.pdf4023 CMOS 三3輸入與非門(mén).pdf4024 CMOS 7級(jí)二進(jìn)制計(jì)數(shù)器.pdf4025 CMOS 三3輸入或非門(mén).pdf40257 CMOS 四2線-1線數(shù)據(jù)選擇器-多路傳輸.pdf4026 CMOS 7段顯示十進(jìn)制計(jì)數(shù)-分頻器.pdf4027 CMOS 帶置位復(fù)位雙J-K主從觸發(fā)器.pdf4028 CMOS BCD- 十進(jìn)制譯碼器.pdf4029 CMOS 可預(yù)制加-減（十-二進(jìn)制）計(jì)數(shù)器.pdf4030 CMOS 四異或門(mén).pdf4031 CMOS 64級(jí)靜態(tài)移位寄存器.pdf4032 CMOS 3位正邏輯串行加法器.pdf4033 CMOS 十進(jìn)制計(jì)數(shù)器-消隱7段顯示.pdf4034 CMOS 8位雙向并、串入-并出寄存器.pdf4035 CMOS 4位并入-并出移位寄存器.pdf4038 CMOS 3位串行負(fù)邏輯加法器.pdf4040 CMOS 12級(jí)二進(jìn)制計(jì)數(shù)-分頻器.pdf4041 CMOS 四原碼-補(bǔ)碼緩沖器.pdf4042 CMOS 四時(shí)鐘控制 D 鎖存器.pdf4043 CMOS 四三態(tài)或非 R-S 鎖存器.pdf4044 CMOS 四三態(tài)與非 R-S 鎖存器.pdf4045 CMOS 21位計(jì)數(shù)器.pdf4046 CMOS PLL 鎖相環(huán)電路.pdf4047 CMOS 單穩(wěn)態(tài)、無(wú)穩(wěn)態(tài)多諧振蕩器.pdf4048 CMOS 8輸入端多功能可擴(kuò)展三態(tài)門(mén).pdf4049 CMOS 六反相緩沖器-轉(zhuǎn)換器.pdf4050 CMOS 六同相緩沖器-轉(zhuǎn)換器.pdf4051 CMOS 8選1雙向模擬開(kāi)關(guān).pdf4051,2,3.pdf4052 CMOS 雙4選1雙向模擬開(kāi)關(guān).pdf4053 CMOS 三2選1雙向模擬開(kāi)關(guān).pdf4054 C
上傳時(shí)間： 2021-11-09
上傳用戶(hù)：kent
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