1079 Chinese remainder theorem (template)

A positive integer K, gives the result of some prime number K Mod, meet the conditions required minimum K. E.g., K% 2 = 1, K% 3 = 2, K% 5 = 3. Eligible minimum of K = 23.

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Entry

Line 1: The number N represents the number 1 and the prime analog input later. (2 <= N <= 10 ) 
of 2 - N + 1 lines, each P is the number 2 and M, separated by a space, P is a prime number, M is the result of K% P. (2 <= P <= 100 , 0 <= K <P)

Export

Output qualified smallest K. All data were less than 10 K ^ 9.

SAMPLE INPUT

3
2 1
3 2
5 3

Sample Output

23 

Basics:

x ≡a1 (v m1) x ≡a2 (v m2) ...... x ≡an (v mn);

Wherein, m1.m2, m3 ....., mn pairwise relatively prime integers

令M=（m1*m2*m3....*mn）；

Mi = M / mi, ti is an equation Mi * ti a solution ≡1 (mod mi) of

Then x has an integer solution, the solution is x = sum 1 ~ n (ai * Mi * ti);

Certify as follows:

Because Mi = m / mi mi other than multiples of, so if all the modulus = [Sigma substituting X (n-~. 1) * AI * Mi Ti , the original equation established

Can be understood as each aiMiti i only contribute to this equation, have no effect on other equations

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll arr[15],brr[15];
 5 int n;
 6 ll X,Y;
 7 ll gdc(ll a,ll b) { return b==0?a:gdc(b,a%b); }
 8 void exgdc(ll a,ll b,ll &X,ll &Y){
 9     if(b==0){
10         X=1,Y=0;return;
11     }
12     exgdc(b,a%b,X,Y);
13     ll temp=X;
14     X=Y;
15     Y=temp-a/b*Y;
16 }
17 
18 
19 int main(){
20     ios::sync_with_stdio(false);
21     cin>>n;
22     ll lcm=1;
23     for(int i=1;i<=n;i++){
24         cin>>arr[i]>>brr[i];
25         lcm*=arr[i];
26     }
27     ll res=0;
28     for(int i=1;i<=n;i++){
29         exgdc(lcm/arr[i],arr[i],X,Y);
30         X=(X+arr[i])%(arr[i]);
31         res=(res+brr[i]*X*lcm/arr[i]);
32     }
33     cout << res%lcm << endl;
34     return 0;
35 }

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