POJ2891StrangeWaytoExpressIntegers(exgcd—解一元线性同余方程组)

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?

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 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.

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

题意：给出k组ai与ri，求满足任意组的 m%ai=ri的m的最小正整数，若不存在输出-1.

题解：额，智商捉鸡(⊙o⊙)… 断断续续看了一天，翻了好几本书，才有点小眉头，终于把书上那些定理证出来了，不过估计很快又会忘了/(ㄒoㄒ)/~~  

这里给出了 m mod(ai)=ri ，那么得到 ai*k+ri=m  ，两边同时 mod (ai)得到  ri mod(ai) ≡ m。  这里给出了k组ai 与 ri，那么我们连立k组 ri mod (ai) ≡ m ，求解即可。 得到 ： 

对于 x≡r1(mod a1)
      x≡r2(mod a2)
相当于解不定方程：x*a1+y*a2=r2-r1

代码如下：

#include
#include
#include
using namespace std;
#define LL long long 

void exgcd(LL a,LL b,LL &d,LL &x,LL &y)
{
	if(!b)
	{
		x=1; y=0;
		d=a;
	}
	else
	{
		exgcd(b,a%b,d,y,x);
		y-=x*(a/b);
	}
}

int main()
{
	LL a1,a2,r1,r2,i,n;
	while(scanf("%lld",&n)!=EOF)
	{
		int flag=1;
		scanf("%lld%lld",&a1,&r1);
		for(i=1;i		{
			scanf("%lld%lld",&a2,&r2);
			LL a=a1,b=a2,d,c=r2-r1,x0,y0;
			exgcd(a,b,d,x0,y0);
			if(c%d!=0)
				flag=0;
			LL s=b/d;
			x0=(x0*(c/d)%s+s)%s;//a*x+b*y==c的最小整数解 
			r1=r1+a1*x0;//迭代r1的值 
			a1=a1*(a2/d);//取a1和a2的公倍数 
		}
		if(!flag)
			printf("-1\n");
		else
			printf("%I64d\n",r1);
	}
	return 0;
}