3293. A Sequence of Numbers

最新推荐文章于 2019-02-20 20:11:34 发布

章鱼鱼的ACM

最新推荐文章于 2019-02-20 20:11:34 发布

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本文链接：https://blog.youkuaiyun.com/u010643814/article/details/37994153

本文探讨了如何通过减少时间复杂度来解决等比数列问题，特别关注于快速找到特定位置的元素。文章提供了一种算法，通过分析等比数列的性质，利用长整型数据类型进行高效计算，并给出具体实现代码。

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注意超时问题，在计算等比数列时注意减少时间复杂度，还有参数的类型均为long long型。

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 ≤ 10⁹, and the other three numbers are in the range [0, 2⁶³). 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>
using namespace std;
const int mod=200907;
int main(){
	long long  N,k,n,m,a,b,c,r;
	cin>>N;
	while(N--){
		cin>>a>>b>>c>>k;
		if(b*2==a+c){
			n=(long long)(b-a);
			r=(a%mod+((k-1)%mod)*n%mod)%mod;
			cout<<r<<endl;
		}
		else{
			r=a;
			m=b/a;
			//cout<<m<<endl;
			k--;
			while(k!=0){
				if(k%2)
					r=(r*m)%mod	;
				k=k/2;
				m=(m*m)%mod;
			}
			cout<<r<<endl;

		}
	}
	return 0;
}