Time Limit: 1000MS | Memory Limit: 131072K | |
Total Submissions: 10907 | Accepted: 3336 |
Description
Choose k different positive integers a1, a2, …, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) can be used to express m.
“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”
Since Elina is new to programming, this problem is too difficult for her. Can you help her?
Input
The input contains multiple test cases. Each test cases consists of some lines.
- Line 1: Contains the integer k.
- Lines 2 ~ k + 1: Each contains a pair of integers ai, ri (1 ≤ i ≤ k).
Output
Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.
Sample Input
2
8 7
11 9
Sample Output
31
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
#define ll long long
#define N 10010 ll exgcd(ll a,ll b,ll& x, ll& y)
{
if(b==)
{
x=;
y=;
return a;
}
ll d=exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
bool solve(ll &m0,ll &a0,ll m,ll a)
{
ll y,x;
ll g=exgcd(m0,m,x,y);
if((a-a0)%g) return ;
x*=(a-a0)/g;
x%=m/g;
a0=(x*m0+a0);
m0*=m/g;
a0%=m0;
if(a0<) a0+=m0;
return ;
}
bool MLES(ll m[],ll r[],ll &m0 ,ll &a0,ll n)
{
m0=;
a0=;
bool flag=;
for(ll i=;i<n;i++)
{
if(!solve(m0,a0,m[i],r[i]))
{
flag=;
break;
}
}
return flag;
}
int main()
{
ll n;
ll m[N],r[N];
while(scanf("%lld",&n)!=EOF)
{
for(ll i=;i<n;i++)
{
scanf("%lld%lld",&m[i],&r[i]);
}
ll m0,a0;
ll flag=MLES(m,r,m0,a0,n);
if(flag) printf("%lld\n",a0);
else printf("-1\n");
}
return ;
}