hdu 2817 A sequence of numbers(判断是等差还是等比数列)

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2817

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

21 2 3 51 2 4 5

 

Sample Output

516

 

Source

2009 Multi-University Training Contest 1 - Host by TJU 

这题的输入是一个数列中的前三个数，然后根据输入判断是等差数列还是等比数列，然后求出第n项。
 
需要注意的地方是对结果取摸：A×B%M = ((A%M)*((B%M)%M)
 
最重要是求等比数列的时候： 求幂乘要用二分来乘，否则会TLE

代码如下：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int a[1000047];// m^n % k__int64 quickpow(__int64 m,__int64 n,__int64 k){    __int64 ans = 1;    while (n > 0)    {if (n & 1) //n%2 == 1ans = (ans*m)%k;n = n >> 1 ;//n /= 2m = (m*m)%k;    }    return ans;} int main(){__int64 n ,t, i,j,a,b,c,k,s,q,d;while(~scanf("%I64d",&n)){while(n--){s = 0;scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&k);if(b-a == c - b){d = b -a;s = ((a%200907)+(((k-1)%200907)*(d%200907))%200907)%200907;}else if(b/a == c/b){q = b/a;s=((a%200907)*(quickpow(q,k-1,200907)))%200907;}printf("%I64d\n",s);}}return 0;}