A sequence of numbers

最新推荐文章于 2020-08-04 20:56:30 发布

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最新推荐文章于 2020-08-04 20:56:30 发布

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博客围绕求解序列第K个数并取模的问题展开。输入包含序列数量N，后续每行有四个整数，前三个是序列前三项，最后一个是K。输出为每个序列第K个数对200907取模的结果，同时提醒运算过程中要取模，否则会出错。

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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

注意点
计算res *= a;和a*=a后（运算过程中），都要记得模m，否则会WA

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

#define m 200907

long long cal(long long a, long long b){
    //二分求幂 a^b
    long long res = 1;

    //31
    while(b){

        //该位为1
        if(b%2 == 1){
            res *= a;    //* 2 * (2^2) * (2^4) * (2^8) * (2^16) = 2^(1+2+4+8+16) = 2^31
            res %= m;
        }
        a *= a;          //2^2 2^4 2^8 2^16 2^32
        a %= m;
        b /= 2;          //15 7 3 1 0
    }
    return res;

}

int main(){
    int N;
    cin >> N;

    long long a,b,c,k;
    long long ans;

    for(int i=0;i<N;i++){
        cin >> a >> b >> c >> k;
        if(b-a == c-b){
            //等差数列
            ans = (a + (k-1) * (b-a))% m;
        }
        else{
            //等比数列
            //q0 * q ^ (k-1)
            //q0 = a, q = b/a
            ans = (a * cal(b/a,k-1)) % m;
        }
        cout << ans << endl;
    }
}