HDU2817(二分幂)

A sequence of numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1830    Accepted Submission(s): 558

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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    注意二分幂的计算，不可直接乘k-3次，也不可不加优化直接递归，注意细节：中间过程也可能产生很大的数
   
   
   
   
   
   
   
   

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


const int mod=200907;

void func1(__int64 b,__int64 c,int k)
{
	k=k%mod;
	__int64 temp=(c-b)%mod;
	__int64 ans=(c+(temp*k)%mod)%mod;
	printf("%I64d\n",ans);
}
	

__int64 binary(__int64 temp,int k)
{
	__int64 sum;
	if(k==1)
		return temp%mod;
	sum=binary(temp,k/2)%mod;
	sum=((sum%mod)*(sum%mod))%mod;
   if(k%2)
	return 	((sum%mod)*(temp%mod))%mod;
	return sum%mod;
}


void func2(__int64 b,__int64 c,int k)
{
	__int64 ans;
	ans=(c%mod*(binary((c/b)%mod,k)%mod))%mod;
	printf("%I64d\n",ans);
}

int main()
{
	__int64 a,b,c;
    int k;
	int t;
	cin>>t;
	while(t--)
	{
		scanf("%I64d%I64d%I64d%d",&a,&b,&c,&k);
		k=k-3;
		if(c-b==b-a)
			func1(b,c,k);
		else
			func2(b,c,k);
	}
	return 0;
}