output(long b,long i) { printf("\n%ld/%ld=809*%ld+%ld",b,i,i,b%i); } main() { long int a,b,i; a=809; for(i=10;i<100;i++) { b=i*a+1; if(b>=1000&&b<=10000&&8*i<100&&9*i>=100) output(b,i); } } |
main() { char *p,s[6];int n; p=s; gets(p); n=0; while(*(p)!='\0') { n=n*8+*p-'0'; p++; } printf("%d",n); } |
main() { long sum=4,s=4; int j; for(j=2;j<=8;j++)/*j is place of number*/ { printf("\n%ld",sum); if(j<=2) s*=7; else s*=8; sum+=s; } printf("\nsum=%ld",sum); } |
#include "stdio.h" #include "math.h" main() { int a,b,c,d; scanf("%d",&a); for(b=3;b<=a/2;b+=2) { for(c=2;c<=sqrt(b);c++) if(b%c==0) break; if(c>sqrt(b)) d=a-b; else break; for(c=2;c<=sqrt(d);c++) if(d%c==0) break; if(c>sqrt(d)) printf("%d=%d+%d\n",a,b,d); } } |
#p#副标题#e#
「程序85」
题目:判断一个素数能被几个9整除
1.程序分析:
2.程序源代码:
main() { long int m9=9,sum=9; int zi,n1=1,c9=1; scanf("%d",&zi); while(n1!=0) { if(!(sum%zi)) n1=0; else { m9=m9*10; sum=sum+m9; c9++; } } printf("%ld,can be divided by %d \"9\"",sum,c9); } |
#include "stdio.h" main() { char a[]="acegikm"; char b[]="bdfhjlnpq"; char c[80],*p; int i=0,j=0,k=0; while(a[i]!='\0'&&b[j]!='\0') { if (a[i] { c[k]=a[i];i++;} else c[k]=b[j++]; k++; } c[k]='\0'; if(a[i]=='\0') p=b+j; else p=a+i; strcat(c,p); puts(c); } |
#include "stdio.h" struct student { int x; char c; } a; main() { a.x=3; a.c='a'; f(a); printf("%d,%c",a.x,a.c); } f(struct student b) { b.x=20; b.c='y'; } |
#p#副标题#e#
「程序88」
题目:读取7个数(1—50)的整数值,每读取一个值,程序打印出该值个数的*。
1.程序分析:
2.程序源代码:
main() { int i,a,n=1; while(n<=7) { do { scanf("%d",&a); }while(a<1||a>50); for(i=1;i<=a;i++) printf("*"); printf("\n"); n++; } getch(); } |
main() { int a,i,aa[4],t; scanf("%d",&a); aa[0]=a%10; aa[1]=a%100/10; aa[2]=a%1000/100; aa[3]=a/1000; for(i=0;i<=3;i++) { aa[i]+=5; aa[i]%=10; } for(i=0;i<=3/2;i++) { t=aa[i]; aa[i]=aa[3-i]; aa[3-i]=t; } for(i=3;i>=0;i--) printf("%d",aa[i]); } |
#include "stdio.h" #define M 5 main() { int a[M]={1,2,3,4,5}; int i,j,t; i=0;j=M-1; while(i {t=*(a+i); *(a+i)=*(a+j); *(a+j)=t; i++;j--; } for(i=0;i printf("%d",*(a+i)); } |