【中国剩余定理/扩展欧几里得定理】AcWing204.表达整数的奇怪方式——题解顺带中国剩余定理模板

AcWing204.表达整数的奇怪方式

题解

模板

根据题目变形

#include <iostream>

using namespace std;

typedef long long LL;

LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if(!b)
    {
        x = 1,  y = 0;
        return a;
    }
    LL d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

int main()
{
    int n;
    cin >> n;
    LL m1, a1, x;
    cin >> m1 >> a1;
    for(int i = 0; i < n - 1; ++i)
    {
        LL m2, a2, k1, k2;
        cin >> m2 >> a2;
        LL d = exgcd(m1, m2, k1, k2);
        if( (a2 - a1) % d  )
        {
            x = -1;
            break;
        }
        k1 *= (a2 - a1) / d;
        k1 = (k1 % (m2 / d) + m2/d) % (m2 / d); //另k1最小使得a最小才能出现最小x
        a1 = k1 * m1 + a1;
        m1 = m1 / d * m2;
    }
    if(x != -1)
        x = (a1 % m1 + m1) % m1;
    cout << x << endl;
    return 0;
}

标签：剩余,题解,定理,a1,k1,a2,m1,m2,LL
来源： https://www.cnblogs.com/czy-algorithm/p/16367782.html