HDU 2817 A sequence of numbers (快速幂)

A sequence of numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6823    Accepted Submission(s): 2073

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

#include<iostream>
#include<cmath>
using namespace std;

typedef long long ll;

int mod = 200907;

ll mod_pow(ll x, ll n)
{
	ll y = 1;
	while(n > 0)
	{
		if(n & 1)    //位运算 判断二进制最后一位是否为1 即是否为奇数
			y = y * x % mod;
		x = x * x % mod;
		n >>= 1;    //按位向右移动 即除以2
	}
	return y;
}

int main()
{
	int n;
	cin >> n;
	while(n--)
	{
		ll a, b, c, k;
		cin >> a >> b >> c >> k;
		if(b - a == c - b)
		{
			ll ans = a % mod + (b - a) * (k - 1) % mod;
			cout << ans % mod << endl;
		}
		else
		{
			ll ans = a * mod_pow((b / a), (k - 1));
			cout << ans % mod << endl;
		}
	}
	return 0;
}