hdu2817 A sequence of numbers

最新推荐文章于 2020-07-07 21:38:00 发布

my1995

最新推荐文章于 2020-07-07 21:38:00 发布

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分类专栏：快速幂文章标签：快速幂

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本文介绍了一种利用快速幂解决特定数列问题的方法。针对已知数列的前三项，快速判断数列类型（等差或等比），并求解第K项的值（取模200907）。提供了完整的C语言代码实现。

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A sequence of numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3267 Accepted Submission(s): 976

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

这是一道快速幂的题目。。。。其实不难，有模版。。吐舌头

#include<stdio.h>
__int64 mod_exp(__int64 a,__int64 b,__int64 c)
{
    __int64 res,t;
    res=1%c;
    t=a%c;
    while(b)
    {
        if(b&1)
            res=res*t%c;
        t=t*t%c;
        b>>=1;
    }
    return res;
}

int main()
{
    __int64 a[3],i,k,d,q,m,n,j;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(i=0;i<3;i++)
        {
            scanf("%I64d",&a[i]);
        }
        scanf("%I64d",&k);
        if((a[0]+a[2])==2*a[1])
        {
            d=a[1]-a[0];
            n=(a[0]+(k-1)*d)%200907;
            printf("%I64d\n",n);
        }
        else
        {
            q=a[1]/a[0];
            m=((a[0]%200907)*mod_exp(q,k-1,200907))%200907;
            printf("%I64d\n",m);
        }
    }
    return 0;
}