中国剩余定理 hdu 3579

HDU 3579 Hello Kiki

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3107 Accepted Submission(s): 1157

Problem Description

One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing "门前大桥下游过一群鸭，快来快来数一数，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

14 57

5 56

19 54 40 24 80

11 2 36 20 76

Sample Output

Case 1: 341

Case 2: 5996

 1 /*看了题解，才知道数据中有只有一组数据，并且整除的数据
 2 if(n==1&&a[1]==0)
 3 {
 4             printf("Case %d: %d
",opt,m[1]);
 5             continue;        
 6 }只有一组数据，并且还整除，中国剩余定理是解决不了的，要特判。
 7 */
 8 #include<iostream> 
 9 using namespace std;
10 #include<cstdio>
11 #define inf (1<<31)-1
12 #define N 10
13 void exgcd(int a,int b,int &x,int &y,int &gcd)
14 {
15     if(b==0)
16     {
17         x=1;y=0;
18         gcd=a;
19         return ;
20     }
21     exgcd(b,a%b,x,y,gcd);
22     int t=x;
23     x=y;
24     y=t-(a/b)*y;
25 }
26 int main()
27 {
28     int T;
29     scanf("%d",&T);
30     int opt=0;
31     while(T--)
32     {
33         ++opt;
34         int n,m[N]={0},a[N]={0};
35         int m1,m2,a1,a2,x,y,gcd;
36         scanf("%d",&n);
37         for(int i=1;i<=n;++i)
38           scanf("%d",&m[i]);
39         for(int i=1;i<=n;++i)
40           scanf("%d",&a[i]);
41         m1=m[1];a1=a[1];
42         if(n==1&&a[1]==0)
43         {
44             printf("Case %d: %d
",opt,m[1]);
45             continue;
46         }
47         bool flag=false;
48         for(int i=2;i<=n;++i)
49         {
50             a2=a[i];m2=m[i];
51             exgcd(m1,m2,x,y,gcd);
52             if((a2-a1)%gcd)
53             {
54                 flag=true;
55                 break;
56             }
57             int t=m2/gcd;
58             x=(x*(a2-a1))/gcd;
59             x=(x%t+t)%t;
60             a1=m1*x+a1;
61             m1=(m1*m2)/gcd;
62             a1=(a1%m1+m1)%m1;
63         }
64         if(flag)
65           printf("Case %d: -1
",opt);
66         else printf("Case %d: %d
",opt,a1);
67     }
68     return 0;
69 }