POJ - 2891 中国剩余定理模板题

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 1e6+10;
typedef long long ll;
typedef pair<ll,ll> pll;
ll a[maxn],b[maxn],m[maxn];

ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}

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

ll inv(ll a,ll p)
{
ll d,x,y;
d = ex_gcd(a,p,x,y);
return d==1?(x%p+p)%p:-1;
}

pll CRT(int n, ll A[], ll B[], ll M[]) {
ll ans = 0, m = 1;
for(int i = 0; i < n; i ++) {
ll a = A[i] * m, b = B[i] - A[i]*ans, d = gcd(M[i], a);
if(b % d != 0)
return pll(0, -1);//答案不存在，返回-1
ll t = b/d * inv(a/d, M[i]/d)%(M[i]/d);
ans = ans + m*t;
m *= M[i]/d;
}
ans = (ans % m + m ) % m;
return pll(ans, m);//返回的x就是答案，m是最后的lcm值
}

int main()
{
int k;
while(~scanf("%d",&k))
{
for(int i=0;i<k;i++)
{
a[i] = 1;
scanf("%I64d %I64d",&m[i],&b[i]);
}
pll ans = CRT(k,a,b,m);
if(ans.second==-1)
printf("-1\n");
else
cout<<ans.first<<endl;
}
}