Poj 2891 Strange Way to Express Integers【crt】

最新推荐文章于 2019-04-08 20:39:24 发布

hope1262946533

最新推荐文章于 2019-04-08 20:39:24 发布

阅读量311

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/liuke19950717/article/details/51292172

本文介绍了一种利用中国剩余定理解决特定整数表达问题的方法。通过给出多个除数和余数组合，找到能够同时满足这些条件的最小非负整数。文章提供了一个C++实现示例，包括扩展欧几里得算法和中国剩余定理的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 12923		Accepted: 4126

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, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) 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 a_i, r_i (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.

Source

POJ Monthly--2006.07.30, Static

题意：

题目说的是，给出若干组除数和余数，求能确定满足所有组的最小的数，如果不存在，输出-1

题解：

中国剩余定理，不是特别懂，无解的情况是基于拓展欧几里得...

/*
http://blog.youkuaiyun.com/liuke19950717
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn=5005;
void extgcd(ll a,ll b,ll &d,ll &x,ll &y)
{
	if(!b)
	{
		d=a;x=1;y=0;
		return;
	}
	extgcd(b,a%b,d,y,x);
	y-=(a/b)*x;
}
ll crt(ll n,ll a[],ll r[])
{
	ll M=a[0],R=r[0],x,y,d;
	for(int i=1;i<n;++i)
	{
		extgcd(M,a[i],d,x,y);
		if((r[i]-R)%d)
		{
			return -1;
		}
		x=(r[i]-R)/d*x%(a[i]/d);
		R+=x*M;
		M=M/d*a[i];
		R%=M;
	}
	return R>0?R:R+M;
}
int main()
{
	ll n;
	while(~scanf("%lld",&n))
	{
		ll a[maxn]={0},r[maxn]={0};
		for(ll i=0;i<n;++i)
		{
			scanf("%lld%lld",&a[i],&r[i]);
		}
		printf("%lld\n",crt(n,a,r));
	}
	return 0;
}