poj 2891 Strange Way to Express Integers (非互质的中国剩余定理)

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 9472		Accepted: 2873

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.

Source

POJ Monthly--2006.07.30, Static

 

参考：http://yzmduncan.iteye.com/blog/1323599/

 

 

直接翻译成代码。

 

//156K    16MS    C++    1362B    2014-06-13 12:36:23
#include<stdio.h>
__int64 gcd(__int64 a,__int64 b)
{
    return b?gcd(b,a%b):a;
}
__int64 extend_euclid(__int64 a,__int64 b,__int64 &x,__int64 &y)
{
    if(b==0){
        x=1;y=0;
        return a;
    }
    __int64 d=extend_euclid(b,a%b,x,y);
    __int64 t=x;
    x=y;
    y=t-a/b*y;
    return d;
}
__int64 inv(__int64 a,__int64 n)
{
    __int64 x,y;
    __int64 t=extend_euclid(a,n,x,y);
    if(t!=1) return -1;
    return (x%n+n)%n;
}
bool merge(__int64 a1,__int64 n1,__int64 a2,__int64 n2,__int64 &a3,__int64 &n3)
{
    __int64 d=gcd(n1,n2);
    __int64 c=a2-a1;
    if(c%d) return false;
    c=(c%n2+n2)%n2;
    c/=d;
    n1/=d;
    n2/=d;
    c*=inv(n1,n2);
    c%=n2;
    c*=n1*d;
    c+=a1;
    n3=n1*n2*d;
    a3=(c%n3+n3)%n3;
    return true;
}
__int64 china_reminder2(int len,__int64 *a,__int64 *n)
{
    __int64 a1=a[0],n1=n[0];
    __int64 a2,n2;
    for(int i=1;i<len;i++){
         __int64 aa,nn;
         a2=a[i],n2=n[i];
         if(!merge(a1,n1,a2,n2,aa,nn)) return -1;
         a1=aa;
         n1=nn;
    }
    return (a1%n1+n1)%n1;
}
int main(void)
{
    int n;
    __int64 a[1005],b[1005];
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
            scanf("%I64d %I64d",&a[i],&b[i]);
        printf("%I64d
",china_reminder2(n,b,a));
    }
    return 0;
}