Hello Kiki

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5489    Accepted Submission(s):
2164

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

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

 1 #include <iostream>
 2 #include<stdio.h>
 3 #include <algorithm>
 4 #include<string.h>
 5 #include<cstring>
 6 #include<math.h>
 7 #define inf 0x3f3f3f3f
 8 #define ll long long
 9 using namespace std;
10
11 int m[15];
12 int r[15];
13 int n,x,y;
14 int gcd;
15
16 int exgcd(int a,int b,int &x,int &y)
17 {
18     if(b==0)
19     {
20         x=1;
21         y=0;
22         return a;
23     }
24     int q=exgcd(b,a%b,y,x);
25     y=y-(a/b)*x;
26     return q;
27 }
28
29 int main()
30 {
31     int t;
32     scanf("%d",&t);
33     for(int cnt=1;cnt<=t;cnt++)
34     {
35         bool flag=true;
36         scanf("%d",&n);
37         for(int i=0;i<n;i++)
38             scanf("%d",&m[i]);
39         for(int i=0;i<n;i++)
40             scanf("%d",&r[i]);
41         int a1=m[0];
42         int r1=r[0];
43         for(int i=1;i<n;i++)
44         {
45             int b1=m[i];
46             int r2=r[i];
47             int d=r2-r1;
48             gcd=exgcd(a1,b1,x,y);
49             if(d%gcd) {flag=false;break;}
50             int multiple=d/gcd;
51             int p=b1/gcd;
52             x=( (x*multiple)%p+p )%p;
53             r1=r1+x*a1;
54             a1=a1*b1/gcd;
55         }
56         if(flag)
57         {
58             if(r1==0) r1=a1+r1;///坑：如果余数是0则加一个最小公倍数
59             printf("Case %d: %d\n",cnt,r1);
60         }
61         else printf("Case %d: -1\n",cnt);
62     }
63     return 0;
64 }

原文地址：https://www.cnblogs.com/shoulinniao/p/10363663.html