【数论 CRT——升级版】 poj 2891 Strange Way to Express Integers

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 25378		Accepted: 8411

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

31

Hint

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

 

这是中国剩余定理的升级版 —— 每个mi不一定两两互质

这样就需要我们通过一下合并两两方程来解决：

 

如合并n%a1=b1,n%a2=b2

即a1*x+b1=a2*y+b2

a1*x-a2*y=b2-b1

设a=a1/t,b=a2/t,c=(b2-b1)/t t=gcd(a,b)

若(b2-b1)%t!=0则无解

用exgcd得到a*x+b*y=c的解x0，

通解x=x0+k*b,k为整数

带入a1*x+b1=n

a1*b*k+a1*x0+b1=n

所以b1=b1+a1*x0，a1=a1*b

//by hzwer

 

代码

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
    return b?gcd(b,a%b):a;
}
void exgcd(ll a,ll b,ll &x,ll &y)
{
    if(!b)
    {
        x=1; y=0;
        return;
    }
    exgcd(b,a%b,x,y);
    ll t=x;
    x=y;
    y=t-(a/b)*y;
}
int main()
{
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    int t;  
    while(scanf("%d",&t)!=EOF)
    {
        int flag=0,n;
        ll a1,a2,b1,b2;
        scanf("%lld%lld",&a1,&b1);
        for(int i=1;i<t;i++)
        {
            scanf("%lld%lld",&a2,&b2);
            if(flag) continue;
            ll a,b,c,d,x,y;
            a=a1,b=a2,c=b2-b1;
            d=gcd(a,b);
            if(c%d)
            {
                printf("-1\n");
                flag=1;
                continue;
            }
            a/=d,b/=d,c/=d;
            exgcd(a,b,x,y);
            x=((c*x)%b+b)%b;
            b1=b1+a1*x;
            a1=a1*b;
        }
        if(!flag) printf("%lld\n",b1);
    }
    return 0;
}