1019. General Palindromic Number (20)

简单的进制转换与回文数的判定。

#include<stdio.h>int main(){//freopen("in.txt","r",stdin);int n, b;scanf("%d %d",&n,&b);int s[100];int k=0;do{s[k++] = n % b;n /= b;}while(n!=0);//s[k++] = n % b;int i, flag = 1;for(i=0; i<k/2; i++) {if(s[i] != s[k-i-1]) {flag = 0;break;}}if(flag)printf("Yes\n");elseprintf("No\n");for(i=k-1; i>0; i--) {printf("%d ",s[i]);}printf("%d\n",s[i]);return 0;}