POJ 2891 中国剩余定理的非互质形式

中国剩余定理的非互质形式

任意n个表达式一对对处理，故只需处理两个表达式。

x = a(mod m)

x = b(mod n)

km+a = b (mod n)

km = (a-b)(mod n)

利用扩展欧几里得算法求出k

k = k0(mod n/(n,m)) = k0 + h*n/(n,m)

x = km+a = k0*m+a+h*n*m/(n,m) = k0*m+a (mod n*m/(n,m))

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <utility>
#include <vector>
#include <queue>
#include <map>
#include <set>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define INF 0x3f3f3f3f
#define LL long long

using namespace std;
LL m0, m1, a0, a1;

LL Exgcd(LL a,LL b,LL &x,LL &y) // ax+by = (a,b)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL x1,y1,x0,y0;
    x0=1; y0=0;
    x1=0; y1=1;
    x=0; y=1;
    LL r=a%b;
    LL q=(a-r)/b;
    while(r)
    {
        x=x0-q*x1; y=y0-q*y1;
        x0=x1; y0=y1;
        x1=x; y1=y;
        a=b; b=r; r=a%b;
        q=(a-r)/b;
    }
    return b;
}

int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        scanf("%lld%lld", &m0, &a0);
        bool flag = 0;
        while(--n)
        {
            scanf("%lld%lld", &m1, &a1);
            LL x, y, tmp;
            LL d = Exgcd(m0, m1, x, y);
            x = (x%(m1/d)+m1/d)%(m1/d);
            tmp = ((a1-a0)%m1+m1)%m1;
            if(tmp%d != 0)
                flag = 1;
            x = x*(tmp/d)%(m1/d);
            a0 = (x*m0+a0+m0/d*m1)%(m0/d*m1);
            m0 = m0/d*m1;
        }
        if(flag)
            printf("-1\n");
        else
            printf("%lld\n", a0);
    }
    return 0;
}