HDU 2817 A sequence of numbers

http://acm.hdu.edu.cn/showproblem.php?pid=2817

 __int64 pow_mod (__int64 a, __int64 n, __int64 m)快速幂取模函数。

                  A sequence of numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4047    Accepted Submission(s): 1243

Problem Description

Xinlv wrote some sequences on the paper a long time ago, they might be arithmetic or geometric sequences. The numbers are not very clear now, and only the first three numbers of each sequence are recognizable. Xinlv wants to know some numbers in these sequences, and he needs your help.

 

 

Input

The first line contains an integer N, indicting that there are N sequences. Each of the following N lines contain four integers. The first three indicating the first three numbers of the sequence, and the last one is K, indicating that we want to know the K-th numbers of the sequence.

You can assume 0 < K <= 10^9, and the other three numbers are in the range [0, 2^63). All the numbers of the sequences are integers. And the sequences are non-decreasing.

 

 

Output

Output one line for each test case, that is, the K-th number module (%) 200907.

 

 

Sample Input

2

1 2 3 5

1 2 4 5

 

 

Sample Output

5

16

 

#include<stdio.h>
#include<string.h>
#define MOD 200907
   __int64 pow_mod (__int64 a, __int64 n, __int64 m)
{
    if(n==0)
         return 1%m;
     if(n==1)
        return a%m;
     __int64 x=pow_mod(a,n/2,m);
     __int64 ans=x*x%m;
    if(n%2==1) ans=ans*a%m;

 return ans;
}


int main()
{
    double a,b,c;
    int t;
    int k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%d",&a,&b,&c,&k);
        if(a+c==2*b)
        {
            __int64 a1=(__int64 )a;
            __int64 d=(__int64 )(b-a);
            int ans=(a1%MOD+((k-1)%MOD)*(d%MOD))%MOD;
            printf("%d\n",ans);
        }
        else
        {
            __int64 a1=(__int64)a;
            __int64 t1=(__int64)(a1%MOD);
            double q1=(b/a);
            __int64 q2=(__int64)q1;
            __int64 q=(__int64)q2;
            __int64 tmp=pow_mod(q,k-1,MOD);
            int ans=(t1*tmp)%MOD;
             printf("%d\n",ans);
        }
    }
    return 0;
}