hdoj 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): 4384    Accepted Submission(s): 1374

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

 题意：给你一个序列的前三位，判断是等差数列还是等比数列，然后求出这个数列的第k项并输出第k项对200907取模

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long 
#define mod 200907
using namespace std;
LL fun(LL a,LL b)
{
	LL ans=1;
	//a=a%mod;
	while(b)
	{
		if(b&1)
		    ans=(a*ans)%mod;
		b/=2;
		a=(a*a)%mod;
	}
	return ans;
}
int main()
{
	int t;
	LL x,y,x1,y1;
	LL a,b,c,k;
	LL ans;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld%lld%lld",&a,&b,&c,&k);
	    if(2*b==a+c)//等差数列 
	    	printf("%lld
",(a+(c-b)*(k-1))%mod);
	    else //等比数列 
	        printf("%lld
",(((fun((c/b),k-1))%mod)*(a%mod))%mod);
	} 
	return 0;
}