POJ 2891 Strange Way to Express Integers（非互质中国剩余定理）

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef __int64 int64;
int64 Mod;

int64 Gcd(int64 a, int64 b)
{
if(b==0)
return a;
return Gcd(b,a%b);
}

int64 EX_Gcd(int64 a, int64 b, int64&x, int64& y)
{
if(b==0)
{
x=1,y=0;
return a;
}
int64 d = EX_Gcd(b,a%b,x,y);
int64 t = x;
x = y;
y = t - a/b*y;
return d;
}

int64 inv(int64 a, int64 n)
{
int64 x,y;
int64 t = EX_Gcd(a,n,x,y);
if(t != 1)
return -1;
return (x%n+n)%n;//返回逆元（最小的正整数解X）
}

bool Merge(int64 a1, int64 n1, int64 a2, int64 n2, int64& a3, int64& n3)
{
int64 d = Gcd(n1,n2);
int64 c = a2-a1;
if(c%d)
return false;
c = (c%n2+n2)%n2;
c /= d;
n1 /= d;
n2 /= d;
c *= inv(n1,n2);
c %= n2;
c *= n1*d;
c += a1;
n3 = n1*n2*d;
a3 = (c%n3+n3)%n3;
return true;
}

int64 China_Reminder(int len, int64* a, int64* n)
{
int64 a1=a[0],n1=n[0];
int64 a2,n2;
for(int i = 1; i < len; i++)
{
int64 aa,nn;
a2 = a[i],n2=n[i];
if(!Merge(a1,n1,a2,n2,aa,nn))
return -1;
a1 = aa;
n1 = nn;
}
Mod = n1;
return (a1%n1+n1)%n1;
}

int64 a[1000],b[1000];
int main()
{
int k;
while(~scanf("%d",&k))
{
for(int i = 0; i < k; i++)
scanf("%I64d %I64d",&a[i],&b[i]);
printf("%I64d\n",China_Reminder(k,b,a));
}
return 0;
}