POJ 2891 中国剩余定理(不互素）

Posted 2020-10-23 liylho

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 17877		Accepted: 6021

Description

 

Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:

 

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, ” 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 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.

Source

POJ Monthly--2006.07.30, Static

题意：n模k个不同的数,得到相应的余数,求最小的n。 
PS：一般的CRT用来解决模数互素的情况，本题不一定模数互素，因此，一般的中国剩余定理模板是不够的。 
代码：

 1 #include "bits/stdc++.h"
 2 #define db double
 3 #define ll long long
 4 //#define vec vector<ll>
 5 #define Mt  vector<vec>
 6 #define ci(x) scanf("%d",&x)
 7 #define cd(x) scanf("%lf",&x)
 8 #define cl(x) scanf("%lld",&x)
 9 #define pi(x) printf("%d\n",x)
10 #define pd(x) printf("%f\n",x)
11 #define pl(x) printf("%lld\n",x)
12 #define inf 0x3f3f3f3f
13 #define rep(i, x, y) for(int i=x;i<=y;i++)
14 const int N   = 1e6 + 5;
15 const int mod = 1e9 + 7;
16 const int MOD = mod - 1;
17 const db  eps = 1e-10;
18 const db  PI  = acos(-1.0);
19 using namespace std;
20 typedef pair<ll,ll> pll;
21 ll a[N],b[N],m[N];
22 ll gcd(ll x,ll y) {return y==0?x:gcd(y,x%y);}
23 ll ex_gcd(ll a,ll b,ll& x,ll& y)
24 {
25     if(b==0){
26         x = 1,y = 0;
27         return a;
28     }
29     ll d = ex_gcd(b,a%b,y,x);
30     y -= a/b*x;
31     return d;
32 }
33 
34 ll inv(ll a,ll p)
35 {
36     ll d,x,y;
37     d = ex_gcd(a,p,x,y);
38     return d==1?(x%p+p)%p:-1;
39 }
40 pll CRT(ll A[], ll B[], ll M[], int n) {//求解A[i]x = B[i] (mod M[i]),总共n个线性方程组
41     ll x = 0, m = 1;
42     for(int i = 0; i < n; i ++) {
43         ll a = A[i] * m, b = B[i] - A[i]*x, d = gcd(M[i], a);
44         if(b % d != 0)  return pll(0, -1);//答案不存在，返回-1
45         ll t = b/d * inv(a/d, M[i]/d)%(M[i]/d);
46         x = x + m*t;
47         m *= M[i]/d;
48     }
49     x = (x % m + m ) % m;
50     return pll(x, m);//返回的x就是答案，m是最后的lcm值
51 }
52 
53 int main()
54 {
55     int k;
56     while(~scanf("%d",&k))
57     {
58         for(int i=0;i<k;i++)
59         {
60             a[i] = 1;
61             scanf("%lld%lld",&m[i],&b[i]);
62         }
63         pll ans = CRT(a,b,m,k);
64         if(ans.second==-1) puts("-1");
65         else    pl(ans.first);
66     }
67 }