http://poj.org/problem?id=2891

Description

Choose k different positive integers a₁, a₂, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) can be used to express m.

“It is easy to calculate the pairs from m

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 a_i, r_i (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

Hint

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

敲模板磨時間、、

 1 #include <algorithm>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 
 6 #define LL long long
 7 const int N(1005);
 8 LL n,a[N],m[N];
 9 
10 LL exgcd(LL a,LL b,LL &x,LL &y)
11 {
12     if(!b)
13     {
14         x=1; y=0;
15         return a;
16     }
17     LL ret=exgcd(b,a%b,x,y),tmp=x;
18     x=y; y=tmp-a/b*y;
19     return ret; 
20 }
21 LL CRT()
22 {
23     LL ret=m[1],tot=a[1];
24     for(int i=2;i<=n;i++)
25     {
26         LL x,y,tt=a[i],c=m[i]-ret;
27         LL gcd=exgcd(tot,tt,x,y);
28         if(c%gcd) return -1;
29         x=x*c/gcd;
30         LL mod=tt/gcd;
31         x=(x%mod+mod)%mod;
32         ret+=x*tot; tot*=mod;
33     }
34     if(!ret) ret+=tot;
35     return ret;
36 }
37 
38 int main()
39 {
40     for(;~scanf("%lld",&n);)
41     {
42         for(int i=1;i<=n;i++)
43             scanf("%lld%lld",a+i,m+i);
44         printf("%lld\n",CRT());
45     }
46     return 0;
47 }

POJ——T 2891 Strange Way to Express Integers