HDU2817 A sequence of numbers【快速模幂】

A sequence of numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6136    Accepted Submission(s): 1976

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

问题链接：HDU2817 A sequence of numbers。

问题简述：（略）

问题分析：

　　这个问题是输入三个数，判断是等差数列还是等比数列，然后按照相应的数列进行计算，再做模除计算。

程序说明：

　　解决这个问题需要注意两点，一是计算模幂（套路）；二是输入的数据需要使用long long类型。

　　计算模幂需要用二分法（快速模幂），以便加快计算速度。

AC的C语言程序如下：

/* HDU2817 A sequence of numbers */

#include 

#define M 200907

// 模幂计算
long long powermod(long long a, long long n, int m)
{
    long long res = 1L;
    while(n) {
        if(n & 1L) {        // n % 2 == 1
            res *= a;
            res %= m;
        }
        a *= a;
        a %= m;
        n >>= 1;
    }
    return res;
}

int main(void)
{
    int n;
    long long a, b, c, si, k;

    scanf("%d", &n);
    while(n--) {
        // 读入数据
        scanf("%lld%lld%lld%lld", &a, &b, &c, &k);

        // 判定数列类型
        if(b-a == c-b) {
            // 按照等差数列计算
            int d = b - a;
            si = (a + (k - 1)*d) % M;
        } else {
            int q = b / a;

            si = (a * powermod(q, k-1, M)) % M;
        }

        // 输出结果
        printf("%lld\n", si);
    }

    return 0;
}