poj-2891-Strange Way to Express Integers【扩展欧几里得】

最新推荐文章于 2021-05-12 00:48:05 发布

Somethingwll

最新推荐文章于 2021-05-12 00:48:05 发布

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分类专栏： gcd&&lcm&&exgcd

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解决一个数学问题，通过中国剩余定理求解特定方程组，找到满足多个同余方程的最小非负整数解。

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题目链接：点击打开链接

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 14774		Accepted: 4842

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

思路：

求解方程组

X%m1=r1

X%m2=r2
....
X%mn=rn

首先看下两个式子的情况

X%m1=r1

X%m2=r2

联立可得

m1*x+m2*y=r2-r1

用ex_gcd求得一个特解x'

得到X=x'*m1+r1

X的通解X'=X+k*LCM(m1,m2)

上式可化为：X'%LCM(m1,m2)=X

到此即完成了两个式子的合并，再将此式子与后边的式子合并，最后的得到的X'即为答案的通解，求最小整数解即可。

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long
using namespace std;
int n;
LL a1,r1,a2,r2,x,y;
LL exgcd(LL a,LL b,LL &x,LL &y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return a;
	}
	LL ans=exgcd(b,a%b,x,y);
	LL t=x;
	x=y;
	y=t-a/b*y;
	return ans;
}
int main()
{
	while(~scanf("%d",&n))
	{
		scanf("%lld%lld",&a1,&r1);
		bool flag=0;
		for(int i=1;i<n;i++)
		{
			scanf("%lld%lld",&a2,&r2);
			if(flag)	continue;
			LL ans=exgcd(a1,a2,x,y); // //a1*x+a2*y=gcd(a1,a2) 
			if((r2-r1)%ans)	flag=1;
			else
			{
				LL tp=a2/ans; // 因为： x=x0+b/gcd(a,b)*k (k=0,1,2...) 
				x=((r2-r1)/ans*x%tp+tp)%tp;
				r1=x*a1+r1;
				a1=a1/ans*a2; // 最小公倍数 
				r1=(r1%a1+a1)%a1;
			}
		}
		if(flag)
			puts("-1");
		else
			printf("%lld\n",r1);
	}
	return 0;
}