HDU 2817 A sequence of numbers 整数快速幂

Posted 2020-06-29

A sequence of numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4550    Accepted Submission(s): 1444

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

 

Recommend

gaojie

 

 

题目大意:给一个数N，代表N组数据，接下来N行，每行前三个数代表等差或等比数列前三个数，求第K个数。。。

 

思路：分两种情况

等差数列：第K个数是a1+(k-1)*d

等比数列：第K个数是a1*(k-1)*q,公比为整数,K<=10^9,q的k次方用到整数快速幂

注意：__int64型

 

#include<iostream>
#define mod 200907
using namespace std;
__int64 mi(__int64 q,__int64 k)
{
    __int64 sum=1;
    while(k)
    {
    if(k&1) sum=(sum*q)%mod;
    q=(q*q)%mod;
    k>>=1;
    }//整数快速幂
    return sum;
}
int main()
{
    int t;
    __int64 a, b, c, k;
    cin>>t;while(t--)
    {
        cin>>a>>b>>c>>k;
        if(c-b==b-a)
        cout<<(a%mod+(k-1)%mod*((c-b)%mod))%mod<<endl;
        else
        cout<<(mi(c/b,--k)%mod*a%mod)%mod<<endl;
    }
    return 0;
}