POJ 2891- Strange Way to Express Integers CRT 除数非互质

**题意：**给你余数和除数求x 注意除数不一定互质 **思路:**不互质的CRT需要的是将两个余数方程合并，需要用到扩展GCD的性质

合并互质求余方程
m1x -+ m2y = r2 - r1 先用exgcd求出特解x0, y0(m1x + m2y = g)
等式两边都乘以c/g 那么可以得解为 x1 = cx0/g，y1 = cy0/g
由性质可知 通解x' = x1 + km2/g，y' = y1 + km1/g 其中k为任意整数
即 x' = cx0/g + km2/g
设 t = m2/g
为保证x是正数 可以将式子化成 x' = (cx0/d % t + t)%t 因为必定存在k使 cx0/g + kt = ( (cx0/g)%t + t)%t
最后合并的取余式为 X = (m1 * x' + r1) (mod lcm(m1, m2) )

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <utility>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <stack>

#include <queue>

#define LL long long

#define ll __int64

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;



const int INF = 0x3f3f3f3f;

const int N = 1e5+2000;



ll gcd(ll a, ll b)

{

    return b?gcd(b, a % b):a;

}



ll exgcd(ll a, ll b, ll &x, ll &y)

{

    ll d = a;

    if(!b)

    {

        x = 1;

        y = 0;

    }

    else

    {

        d = exgcd(b, a % b, y, x);

        y -= (a / b) * x;

    }

    return d;

}






int main()

{

    ll n;

    while(~scanf("%I64d", &n))

    {

        ll x, y;

        ll m, r;

        ll mt, rt;

        int flag = 0;

        scanf("%I64d%I64d", &m, &r);

        for(int i = 0; i < n - 1; i++)

        {

            scanf("%I64d%I64d", &mt, &rt);

            ll c = rt - r;

            ll g = gcd(m, mt);

            if(c % g!=0)

            {

                flag = 1;

            }

            else

            {

                exgcd(m, mt, x, y);

                ll t = mt / g;

                x = (c / g * x % t + t) % t;

                r = m * x + r;

                m = m /gcd(m, mt) * mt;

            }

        }

        if(flag == 1)

            printf("-1\n");

        else printf("%I64d\n", r);

    }

    return 0;

}