九度OJ 1442 A sequence of numbers_a sequence of numbers 九度教程-优快云博客

博客详细解析了九度在线判题系统（OJ）中的第1442道题目，内容涉及数列的特定规律求解。适合考研及准备机试的读者进行算法训练。

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题目地址：http://ac.jobdu.com/problem.php?pid=1442

题目描述：

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.

输入：

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.

输出：

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

样例输入：

2
1 2 3 5
1 2 4 5

样例输出：

5
16

#include <stdio.h>
 
#define M 200907
 
long long fun1(long long data[], int k){
    long long p = data[1] - data[0];
    return ((data[0] % M) + (((k-1) % M) * (p % M)) % M) % M;
}
 
long long fun2(long long data[], int k){
    long long p = data[1] / data[0];
    long long ans = data[0];
    --k;
    while (k != 0){
        if (k % 2 == 1){
            ans = (ans * p) % M;
        }
        k /= 2;
        p = (p * p) % M;
    }
    return ans;
}
 
long long KthNumber(long long data[], int k){
    int flag;
 
    if ((data[1] - data[0]) == (data[2] - data[1]))
        flag = 0;
    else
        flag = 1;
    if (flag == 0){
        return fun1(data, k);
    }
    else{
        return fun2(data, k);
    }
}
 
int main(void){
    int n;
    long long data[3];
    int k;
    int i;
 
    while (scanf ("%d", &n) != EOF){
        while (n-- != 0){
            for (i=0; i<3; ++i){
                scanf ("%lld", &data[i]);
            }
            scanf ("%d", &k);
            printf ("%d\n", KthNumber (data, k));
        }
    }
 
    return 0;
}