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http://acm.hdu.edu.cn/showproblem.php?pid=3579

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4208 Accepted Submission(s): 1617

Problem Description One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing "門前大橋下遊過一群鴨，快來快來 數一數，二四六七八". And then the cashier put the counted coins back morosely and count again...
Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
One day Kiki‘s father found her note and he wanted to know how much coins Kiki was counting.

Input The first line is T indicating the number of test cases.
Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi

Output For each case output the least positive integer X which Kiki was counting in the sample output format. If there is no solution then output -1.

Sample Input 2 2 14 57 5 56 5 19 54 40 24 80 11 2 36 20 76

Sample Output Case 1: 341 Case 2: 5996

Author digiter (Special Thanks echo)

Source 2010 ACM-ICPC Multi-University Training Contest（14）——Host by BJTU 我的媽呀 CRT 還不互質 shit

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

HDU——T 3579 Hello Kiki