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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30304 Accepted Submission(s):
16811
#include<stdio.h> #include<string.h> int main() { int n,m,j,i,s,y,l2; char b[100]; while(scanf("%d %d",&n,&m)!=EOF) { y=0;l2=0;j=1; s=0; while(n!=0) { if(n<0) //考虑n为负数的情况 { s=n; n=-n; } y=n%m; //求每次短除m后的余数 n=n/m; //求每次短除m后的商 if(y>=10) { l2=j; //记录字符串长度 b[j]=y+55;//将数字转换为字符 j++; } else { l2=j; b[j]=y+48; j++; } } if(s<0) printf("-"); for(i=l2;i>=1;i--) { printf("%c",b[i]); } printf("\n"); } return 0; }
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原文地址:http://www.cnblogs.com/tonghao/p/4457989.html