A sequence of numbers_the sequence of numbers csdn-优快云博客

本文链接：https://blog.youkuaiyun.com/liuxinglon/article/details/20244675

本文介绍了一个基于等差和等比序列的问题求解方法，并提供了一段C语言代码实现。该方法能够根据序列的前三个数字及指定位置，计算出序列中对应项的值并返回其模200907的结果。

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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

Source

2009 Multi-University Training Contest 1 - Host by TJU

#include<stdio.h>
unsigned __int64 pow(__int64 a,__int64 b)
{
    unsigned __int64 c;
    c=1;
    while(b)
    {
        if(b%2==1)
        {
            c*=a;
            c%=200907;
        }
        
        a=((a%200907)*(a%200907))%200907;
        b/=2;
    
    }

    return c;
}

int main()
{
    int n;
    unsigned __int64 a,b,c,d,e;

    
    scanf("%d",&n);
    while(n--)
    {
        scanf("%I64d %I64d %I64d %I64d",&a,&b,&c,&d);

        
        if((c-b)==(b-a))//等差数列
        {
                 e=(a%200907+(((d-1)%200907)*((b-a)%200907))%200907)%200907;    //     等差数列公式 a[n]=a[1]+(n-1)*d;
        }

        else
        {
                  
            
            e=(a%200907*pow(b/a%200907,(d-1))%200907)%200907;    //     等比数列公式  a[n]=a[1]*q^(n-1);

        }

        printf("%I64d\n",e);
    }
    return 0;
}