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+;
 ll n,m[N],c[N];
 void exgcd(ll a,ll b,ll& x,ll& y,ll& g) {
     if(!b)x=,y=,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 ;
     y1*=(c1-c2)/g,c1-=y1%(m2/g)*m1,m1*=m2/g,c1=(c1+m1)%m1;
     return ;
 }
 ll solve() {
     ll C=,M=;
     for(ll i=; i<n; ++i)if(!CRT(C,M,c[i],m[i]))return -;
     return C;
 }

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