【POJ2891】Strange Way to Express Integers——中国剩余定理（非互质）

Strange Way to Express Integers

Time Limit: 1000MS	Memory Limit: 131072K

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

题意：取同余方程组的解。

• 分析：
  
  • 中国剩余定理：中国剩余定理是中国古代求解一次同余方程组的方法，是数论中的一个重要定理。
  • 设 m1,m2,m3,⋯,mk是两两互质的正整数，即GCD(mi,mj)=1(i!=ji,j=1,2,3,⋯,k)。同余方程组为：x=a1(modm1)x=a2(modm2)x=a3(modm3)⋮x=ak(modmk)即在[m1,m2,m3,⋯mk]的意义下，存在x满足x=ai(modmi)则x=∑i=1k(ai×Mi×M′i)(modM)M=∏i=1kmiMi=MmiM′i是在Mi模mi时的逆元。
  • 中国剩余定理（非互质）：中国剩余定理要求的是模数两两互质，而在非互质的时候可以采用合并方程的思想。
  • x=a1(modm1)x=a2(modm2)即：x=m1k1+a1x=m2k2+a2m1k1+a1=m2k2+a2m1k1=(a2−a1)+m2k2m1k1=(a2−a1)(modm2)设d=GCD(m1,m2),c=(a2−a1)则：m1k1d=cd(modm2d)→k1=cd×(m1d)′(modm2d)令K=cd×m1d′,k1=y×m2d+K带入（1）式得：x=m1(ym2d+K)+a1=ym1m2d+Km1+a1即x=a(modm)a=Km1+a1m=m1m2d这样将两个式子合为一个式子，最后将k个式子合为x=a(modm)的一个式子求解。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <string>
using namespace std;

typedef unsigned long long ULL;

typedef long long LL;

typedef struct node
{
    LL a;

    LL b;
}Node;

Node s[11000];

int n;

LL EXGCD(LL a,LL b,LL &x,LL &y) //扩展欧几里得
{
    if(b==0)
    {
        x = 1;

        y = 0;

        return a;
    }
    else
    {
        LL d = EXGCD(b,a%b,y,x);

        y-=x*(a/b);

        return d;
    }
}

LL GCD(LL a,LL b)
{
    return b==0?a:GCD(b,a%b);
}


LL LCM(LL a, LL b)
{
    return a*b/GCD(a,b);
}

LL Inv(LL a,LL b)//求逆元
{
    LL x,y;

    LL d = EXGCD(a,b,x,y);

    if(d != 1)
    {
        return -1;
    }
    else
    {
        return (x%b+b)%b;
    }
}

bool Union(LL a1,LL b1,LL a2, LL b2,LL &a3,LL &b3)
{
    LL d = GCD(a1,a2);

    LL b = (b2-b1);

    if(b%d) return false;

    b = (b%a2+a2)%a2;

    b = (b/d);

    LL c = Inv(a1/d,a2/d);

    b3 = b*c;

    b3%=(a2/d);

    b3 *=a1;

    b3+=b1;

    a3 = a1*a2/d;

    b3 = (b3%a3+a3)%a3;

    return true;


}
LL China()
{
    LL aa = s[0].a,bb = s[0].b ;

    for(int i  = 1; i < n ; i++)
    {
        LL st,bt;


        if(Union(aa,bb,s[i].a,s[i].b,st,bt))
        {
            aa = st;

            bb= bt;
        }
        else
        {
            return -1;
        }
    }

    return (bb%aa+aa%aa);
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i = 0 ; i < n; i++)
        {
            scanf("%lld %lld",&s[i].a,&s[i].b);
        }

        printf("%lld\n",China());
    }

    return 0;
}