POJ 2305：Basic remains 进制转换

Basic remains

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5221 Accepted: 2203

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 10110 123456789123456789123456789 10000

Sample Output

10789

给出一个base进制的数，base的范围好在在2到10之间。然后给出在base进制下的p与m，求在base进制下的p%m。

来回的进制转换。

代码：

#include <iostream>  #include <algorithm>  #include <cmath>  #include <vector>  #include <string>  #include <cstring>  #pragma warning(disable:4996)  using namespace std;int base;string num;string m;int change(string n){int i;int sum = 0;int len = n.length();for (i = 0; i < len; i++){sum = sum*base + n[i] - '0';}return sum;}void res(int mod){int i, n, a[500];int len = num.length();int res = 0;for (i = 0; i < len; i++){res = (res*base + num[i] - '0') % mod;}//转换成base进制n = 0;if (res != 0){while (res != 0){a[n++] = res % base;res = res / base;}for (i = n - 1; i >= 0; i--){printf("%d", a[i]);}}else{printf("0");}printf("\n");}int main(){//freopen("i.txt", "r", stdin);//freopen("o.txt", "w", stdout);int mod;while (cin >> base){if (base == 0)break;cin >> num >> m;mod = change(m);res(mod);}//system("pause");return 0;}