POJ 2305 Basic remains（进制转换）

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[1005],b[10];
long long c,e;//开int会WA
int len1,len2;
int f[10];
int change1()//把b转换成十进制//对的
{
    int d=0,i;
    for(i=0;i<len2;i++)
    {
        d=d*c+b[i]-'0';
    }
    return d;
}
void change2()//把十进制e转化成目标进制输出
{
    int i;
    for(i=0;i<10;i++)
    {
        if(e<c)break;
        f[i]=e%c;
        e/=c;
    }
    f[i]=e;
    int len3=i+1;
    for(i=len3-1;i>=0;i--)
    {
        if(i==len3-1&&f[i]==0&&len3!=1)
            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=0;
        for(i=0;i<len1;i++)
        {
            e=e*c+a[i]-'0';
            e=e%d;
        }
        //cout<<e<<endl;
        change2();
    }
    return 0;
}