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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