POJ - 2891 Strange Way to Express Integers (扩展中国剩余定理)

题目链接

扩展CRT模板题，原理及证明见传送门(引用)

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll N=1e5+10;
ll n,m[N],c[N];
void exgcd(ll a,ll b,ll& x,ll& y,ll& g) {
    if(!b)x=1,y=0,g=a;
    else exgcd(b,a%b,y,x,g),y-=x*(a/b);
}
bool CRT(ll& c1,ll& m1,ll c2,ll m2) {
    ll y1,y2,g;
    exgcd(m1,m2,y1,y2,g);
    if((c1-c2)%g)return 0;
    y1*=(c1-c2)/g,c1-=y1%(m2/g)*m1,m1*=m2/g,c1=(c1+m1)%m1;
    return 1;
}
ll solve() {
    ll C=0,M=1;
    for(ll i=0; i<n; ++i)if(!CRT(C,M,c[i],m[i]))return -1;
    return C;
}

int main() {
    while(scanf("%lld",&n)==1) {
        for(ll i=0; i<n; ++i)scanf("%lld%lld",&m[i],&c[i]);
        printf("%lld\n",solve());
    }
    return 0;
}