POJ 2305：Basic remains 进制转换

Basic remains

Time Limit: 1000MS		Memory Limit: 65536K
Total 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 101

10 123456789123456789123456789 1000

0

Sample Output

10

789

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;

}

巴特西

POJ 2305：Basic remains 进制转换

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