剩余定理（孙子定理）

原创已于 2025-11-21 22:37:59 修改 · 2.3k 阅读

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#算法

于 2021-04-16 15:37:44 首次发布

本文介绍了一种使用中国剩余定理解决在线编程挑战（OJ）中 Strange Way to Express Integers 问题的方法。代码示例展示了如何通过求解非负整数的余数对来找到原始数值，同时提供了输入输出样例和代码实现，涉及的主要算法包括中国剩余定理和欧几里得算法。

一，狭义剩余定理

二，OJ实战

POJ 2891 Strange Way to Express Integers

三，广义剩余定理

1，分解方程

2，合并方程

一，狭义剩余定理

二，OJ实战

POJ 2891 Strange Way to Express Integers

题目：

Description

Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:

Choose k different positive integers a1, a2, …, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) can be used to express m.

“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”

Since Elina is new to programming, this problem is too difficult for her. Can you help her?

Input

The input contains multiple test cases. Each test cases consists of some lines.

Line 1: Contains the integer k.
Lines 2 ~ k + 1: Each contains a pair of integers ai, ri (1 ≤ i ≤ k).
Output

Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.

Sample Input

2
8 7
11 9

Sample Output

Hint

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

代码：

#include <iostream>
using namespace std;
#define l long long
 
l x, y;
 
l gcd(l a, l b)
{
	if (a == 0 || b == 0)
	{
		x = (b == 0);
		y = (a == 0);
		return a + b;
	}
	l r;
	if (a < 0)
	{
		r = gcd(-a, b);
		x *= -1;
		return r;
	}
	if (b < 0)
	{
		r = gcd(a, -b);
		y *= -1;
		return r;
	}
	if (a >= b)r = gcd(a%b, b);
	else r = gcd(a, b%a);
	y -= a / b*x;
	x -= b / a*y;
	return r;
}
 
l cheng(l a, l b, l p)
{
	if (b == 0)return 0;
	if (b < 0)return cheng(a, -b, p)*-1;
	l c = cheng(a, b / 2, p);
	c = c + c;
	if (b % 2)c += a;
	return c%p;
}
 
int main()
{
	l t, a1, r1, a2, r2;
	while (cin >> t)
	{
		a1 = 1, r1 = 0;
		while (t--)
		{
			cin >> a2 >> r2;
			if (r1 >= 0)
			{
				l g = gcd(a1, a2);
				if ((r2 - r1) % g)r1 = -1;
				else
				{
					l a3 = a1 / g*a2;
					r1 += cheng(cheng((r2 - r1) / g, x, a3), a1, a3);
					r1 = (r1%a3 + a3) % a3;
					a1 = a3;
				}
			}			
		}
		cout << r1 << endl;
	}	
	return 0;
}