HDUOJ---hello Kiki

Hello Kiki

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1717    Accepted Submission(s): 599

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

 

Author

digiter (Special Thanks echo)

 

Source

2010 ACM-ICPC Multi-University Training Contest（14）——Host by BJTU

 

Recommend

zhouzeyong

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

中国剩余定理：

该题典型的同余方程组X=amod(M)求解，需要注意的是，题目要求最小的的整数解，所以如果解为0是，他们的最小解为他们的最小公倍数..lcm

代码：

 #include<iostream>
 #include<cstdio>
 #define LL _int64   //long long
 using namespace std;
   LL x,y,q;
  LL gcd(LL a,LL b)
  {
      if(b==)
          return gcd(b,a%b);
      else
          return a;
  }

  void  exgcd( LL a, LL b)
   {
       if(b==)
       x=,y=,q=a;
       else
       {
           exgcd(b,a%b);
           LL temp=x;
          x=y,y=temp-a/b*y;
       }
   }

  int main()
  {
      int ncase,n,i,j;
      LL lcm,aa[],rr[];
      bool  ifhave;
     // freopen("test.in","r",stdin);
      //freopen("test.out","w",stdout);
      scanf("%d",&ncase);
      for(j=;j<=ncase;j++)
      {
        scanf("%d",&n);
        lcm=;
        ifhave=true;
        for(i=;i<n;i++)
        {
            scanf("%I64d",&aa[i]);
            lcm=lcm/gcd(lcm,aa[i])*aa[i];
        }
        for(i=;i<n;i++)
            scanf("%I64d",&rr[i]);
        for(i=;i<n;i++)
        {
           exgcd(aa[],aa[i]);
           if((rr[i]-rr[])%q)
           {
             ifhave=false;
             break;
           }
           int t=aa[i]/q;
            x=(x*((rr[i]-rr[])/q)%t+t)%t;
            rr[]+=x*aa[];
            aa[]*=(aa[i]/q);
        }
        printf("Case %d: ",j);
        if(!ifhave)
        {
            printf("-1\n");
        }
        else
        {
            if(rr[]!=)
            printf("%I64d\n",rr[]);
            else
             printf("%I64d\n",lcm);
        }
      }
   return ;
  }

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