题目链接：http://poj.org/problem?id=2305
| ime Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5326 | Accepted: 2267 |
Description
Given a base b and two non-negative base b integers p and m, compute p mod m and print the result as a base b integer. p mod m is defined as the smallest non-negative integer k such that p = a*m + k for some integer a.
Input
Input consists of a number of cases. Each case is represented by a line containing three unsigned integers. The first, b, is a decimal number between 2 and 10. The second, p, contains up to 1000 digits between 0 and b-1. The third, m, contains up to 9 digits between 0 and b-1. The last case is followed by a line containing 0.
Output
For each test case, print a line giving p mod m as a base-b integer.
Sample Input
2 1100 101
10 123456789123456789123456789 1000
0
Sample Output
10
789
#include<stdio.h>
#include<iostream>
#include<cstring>
#include<limits.h>
#include<queue>
using namespace std;
char a[],b[];
long long c,e;//开int会WA
int len1,len2;
int f[];
int change1()//把b转换成十进制//对的
{
int d=,i;
for(i=;i<len2;i++)
{
d=d*c+b[i]-'';
}
return d;
}
void change2()//把十进制e转化成目标进制输出
{
int i;
for(i=;i<;i++)
{
if(e<c)break;
f[i]=e%c;
e/=c;
}
f[i]=e;
int len3=i+;
for(i=len3-;i>=;i--)
{
if(i==len3-&&f[i]==&&len3!=)
continue;
else
cout<<f[i];
}
cout<<endl;
}
int main()
{
int i;
while(cin>>c&&c)
{
cin>>a>>b;
len1=strlen(a);
len2=strlen(b);
int d=change1();
//cout<<d<<endl;
e=;
for(i=;i<len1;i++)
{
e=e*c+a[i]-'';
e=e%d;
}
//cout<<e<<endl;
change2();
}
return ;
}