POJ2891 Strange Way to Express Integers [中国剩余定理]
不互质情况的模板题
注意多组数据不要一发现不合法就退出
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; inline ll read(){ char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } ll n,a1,m1,a2,m2; void exgcd(ll a,ll b,ll &d,ll &x,ll &y){ if(b==0) d=a,x=1,y=0; else exgcd(b,a%b,d,y,x),y-=(a/b)*x; } int main(){ freopen("in","r",stdin); while(scanf("%lld",&n)!=EOF){ int flag=0; m1=read();a1=read();n--; while(n--){ m2=read();a2=read(); if(flag) continue; ll d,t1,t2; exgcd(m1,m2,d,t1,t2); if((a2-a1)%d){flag=1;continue;} t1*=(a2-a1)/d; m2/=d; t1=(t1%m2+m2)%m2; a1=a1+m1*t1; m1*=m2; } if(flag) puts("-1"); else printf("%lld\n",a1); } }
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