1、.1、用篩選法求100之內(nèi)的素?cái)?shù)。#include <stdio.h>void main() int num100,i,j; for (i=0;i<100;i+) numi=1; for (i=2;i<=10;i+) for (j=2;i*j<=100;j+) numi*j-1=0; printf("0至100內(nèi)素?cái)?shù)有:n"); for (i=j=0;i<100;i+) if (numi=1) printf("%-4d",i+1);if (+j%4=0) printf("n");2、用選擇法對(duì)10個(gè)

2、整數(shù)排序。#include <stdio.h>void main() int num10,n,i,j,t,k; printf("請(qǐng)輸入十個(gè)整數(shù):"); for (n=0;n<10;n+) scanf("%d",&numn); for (i=0;i<9;i+) k=i; for (j=i+1;j<10;j+) if (numk<numj) k=j; if (k!=i) t=numi;numi=numk;numk=t; printf("從大到小排序?yàn)?"); for (n=0;n<10;n

3、+) printf("%d ",numn);3、求一個(gè)3*3的整型二維數(shù)組對(duì)角線(xiàn)元素之和。#include <stdio.h>void main() int num33,i,j; printf("輸入二維數(shù)組:n"); for (i=0;i<3;i+) for (j=0;j<3;j+) scanf("%d",&numij); printf("兩對(duì)角線(xiàn)和分別為%d和%dn",num00+num11+num22,num02+num11+num20);5、將一個(gè)數(shù)組中的值按逆序重新存放。#

4、include <stdio.h>#include <stdlib.h>#include <time.h>void main() int num100,n,i,temp; srand(time(0); n=rand()%100+1; printf("生成數(shù)組長(zhǎng)度為%dn生成的數(shù)組:n",n); for (i=0;i<n;i+) numi=rand()%100;printf("%-3d",numi); printf("n數(shù)組逆序存放后排列為:n"); for (i=0;i<n/2;i+)

5、temp=numi;numi=numn-1-i;numn-1-i=temp; for (i=0;i<n;i+) printf("%-3d",numi); printf("n");6、輸出以下的楊輝三角形（要求輸出10行）。11 11 2 11 3 3 1 #include <stdio.h>void main() int num1010,i,j; for (i=0;i<10;i+) for (j=0;j<10;j+) if (j-1<0|j=i) numij=1; else numij=numi-1j-1+numi-1

6、j; printf("%-4d",numij); if (j=i) break; printf("n"); 7、輸出“魔方陣”。所謂魔方陣是指這樣的方陣，它的每一行、每一列和對(duì)角線(xiàn)之和均相等。例如，三階魔方陣為8 1 6 3 5 7 4 9 2#include <stdio.h>void main() int n,i,j,k; printf("輸入奇數(shù)階魔方陣的階數(shù):"); scanf("%d",&n); k=(5*n-1)/2; for(i=0;i<n;+i) for (j=0;j<

7、;n;j+) printf("%-4d",(k-i+j)%n*n+(k-i-j-1)%n+1); printf("n"); 9、有15個(gè)數(shù)按由大到小順序存放在一個(gè)數(shù)組中，輸入一個(gè)數(shù)，要求用折半查找法找出該數(shù)是數(shù)組中第幾個(gè)元素的值。如果該數(shù)不在數(shù)組中，則輸出“無(wú)此數(shù)”。#include <stdio.h>#include <stdlib.h>#include <time.h>void main() int num15,n,t=0,left,right,middle,key; srand(time(0); printf(&

8、quot;數(shù)組為n"); for (n=14;n>=0;n-) t=numn=rand()%10+(t+1); for (n=0;n<15;n+) printf("%-4d",numn); printf("n請(qǐng)輸入要查找的數(shù):"); scanf("%d",&key); left=0; right=14; while (left<=right) middle=(left+right)/2; if (key=nummiddle) t=-1;break; if (key<nummiddle) lef

9、t=middle+1; else right=middle-1; if (t=-1) printf("該數(shù)是數(shù)組中第%d個(gè)元素的值n",middle+1); else printf("無(wú)此數(shù)n");11、輸出以下圖案：* * * * * * * * * * * * * * * * * * * * * * * *#include <stdio.h>void main() char asterisk="* * * * *",space=" " int a,b; for (a=0;a<5;a+) for

10、 (b=0;b<a;b+) printf("%s",space); printf("%sn",asterisk); 12、有一行電文，已按下面規(guī)律譯成密碼：第1個(gè)字母變成第26個(gè)字母，第i個(gè)字母變成第（26-i+1）個(gè)字母。非字母字符不變。要求編程序?qū)⒚艽a譯回原文，并輸出密碼和原文。#include <stdio.h>#include <ctype.h>void main() char ch,code100; int n=0; printf("請(qǐng)輸入密碼:"); while(ch=getchar()!=&

11、#39;n') if (n=0) printf("密碼n"); printf("%c",ch); if (isupper(ch) coden+=(26-ch+65+64); if (islower(ch) coden+=(26-ch+97+96); if (!isalpha(ch) coden+=ch; coden='0' printf("n原文n%sn",code);13、編一個(gè)程序，將兩個(gè)字符串連接超來(lái)，不要用strcat函數(shù)。#include <stdio.h>#include <str

12、ing.h>void main() char a2100; puts("輸入字符串:"); gets(a0); gets(a1); int i=0; while(a0i!='0') i+; strcpy(&a0i,a1); puts("連接后字符串:"); puts(a0);14、編一個(gè)程序，將兩個(gè)字符串s1和s2比較，若s1>s2，輸出一個(gè)正數(shù)；若s1=s2，輸出0；若s1<s2,輸出一個(gè)負(fù)數(shù)。不要用strcpy函數(shù)。兩個(gè)字符串用gets函數(shù)讀入。輸出的正數(shù)或負(fù)數(shù)的絕對(duì)值應(yīng)是相比較的兩個(gè)字符串相應(yīng)字符的ASCI

13、I碼的差值。#include <stdio.h>void main() char a2100; puts("輸入字符串:"); gets(a0); gets(a1); for (int i=0,n=0;a0i!='0'&&a1i!='0'i+) if (a0i!=a1i) n=a0i-a1i;break; printf("%dn",n);15、編寫(xiě)一個(gè)程序，將字符數(shù)組s2中的全部字符復(fù)制到字符數(shù)組s1中。不用strcpy函數(shù)。復(fù)制時(shí)，0后面的字符不復(fù)制。#include <stdio.h&

14、gt;#include <string.h>void main() char a2100; puts("輸入字符串:"); gets(a0); gets(a1); a00='0' strcat(a0,a1); puts("復(fù)制后字符串:"); puts(a0);從1累加到100 #include <stdio.h>main() int i,sum=0; for(i=1;i<=100;i+) sum=sum+i; printf("sum=%dn",sum); 求1！+2!+n!:#inclu

15、de<stdio.h>Void main() double fac, sum=0；Int i, n;fac=1;i=1;scanf(“%d”,&n);while(i<=n)fac=fac*I;i+;sum=sum+fac;printf(“%lf”,sum);求1-3+5-7+。+101的值#include<stdio.h>#include<math.h>Void main()int a,I,j,sum=0;i=1;j=2;while(1<=101)sum=sum+i*power(-1,j);i=i+2;j+;printf(“%dn”,su

16、m);求1*2*3*100#include <stdio.h>main()int i ,sum;i=1, sum=1;while(i<101)i+;sum=sum*i;printf("sum=%dn",sum);水仙花數(shù)：#include <stdio.h>main() int a,b,c,num; for(a=1;a<10;a+) for(b=0;b<10;b+) for(c=0;c<10;c+) if(a*100+b*10+c=(num=a*a*a+b*b*b+c

17、*c*c) printf("%d",num); 從1到1000的素?cái)?shù)：#include "stdio.h"void main()int f(int n);int i,k=0;for(i=1;i<=1000;i+)if(!f(i)printf("%5d",i);k+;if(k%15=0)printf("n");getch();int f(int n)int j;for(j=2;j<=n-1;j+)if(n%j=0)break;if(j>=n)return 0;elsereturn 1;最大公約數(shù)和最小公倍數(shù)：#include<stdio.h>Void main() int m,n,I,j;Scanf(“m=%d,n=%d”,&m,&n); For(i=m;i>=1;i-)If(m