【模板】【POJ2891】扩展中国剩余定理

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
const int N=1005;

int k;
LL c
,m
;
bool flag;

LL gcd(LL a,LL b)
{
return b?gcd(b,a%b):a;
}

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

LL inv(LL a,LL mod)
{
LL x,y;
exgcd(a,mod,x,y);
return (x%mod+mod)%mod;
}

LL exCRT(int k)
{
LL m1,m2,c1,c2,g;
for(int i=2;i<=k;i++)
{
m1=m[i-1],m2=m[i];
c1=c[i-1],c2=c[i];
g=gcd(m1,m2);
if((c2-c1)%g) {flag=0;return -1;}
m[i]=m1*m2/g;
c[i]=inv(m1/g,m2/g)*((c2-c1)/g)%(m2/g)*m1+c1;
c[i]=(c[i]%m[i]+m[i])%m[i];
}
return c[k];
}

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