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[],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 ;
 }