A sequence of numbers（快速求冪），numbers求冪

A sequence of numbers（快速求冪），numbers求冪

題目描述

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.

輸入要求

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.

輸出要求

Output one line for each test case, that is, the K-th number module (%) 200907.

假如輸入

21 2 3 51 2 4 5

應當輸出

516

題目大意很簡單，就是給你一個序列的前三項，該序列不是等差就是等比，讓你求第K項餘200907。

快速求冪實現代碼為

int fastpow(int a,int b){int r=1,base=a;while(b!=0){if(b&1)r*=base;base*=base;b>>=1;}return r;}

本題ac代碼

#include<iostream>#include<algorithm>#include <vector>#include<string.h>#include<string>#include <cstring>#include<ctype.h>#include<math.h>#include <queue>#include<map>using namespace std;int MOD=200907;long long fastpow(long long q,long long n){long long r=1;while(n){if(n&1)r=r%MOD*(q%MOD)%MOD;q=q%MOD*(q%MOD)%MOD;n>>=1;}return r;}void solve();int main(){solve();return 0;}void solve(){   long long t,a1,a2,a3,n,d,ans,q;cin>>t;while(t--){cin>>a1>>a2>>a3>>n;if(a2*2==a1+a3){d=a2-a1;ans=(a1+(n-1)*d%MOD)%MOD;}else{q=a2/a1;ans=a1*fastpow(q,n-1)%MOD;}cout<<ans<<endl;}}