九度OJ-1442：A sequence of numbers

　　一道大数据处理的题目。因为数据过大故要注意两点：取模＋二分求幂。

　　二分求幂就不细说了。取模用到了如下两个公式：

　　①(a*b)%c=((a%c)*(b%c))%c

　　②(a+b)%c=((a%c)+(b%c))%c

题目描述：

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.

输入：

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.

输出：

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

样例输入：

21 2 3 51 2 4 5

样例输出：

516

#include <iostream> using namespace std;long long pow(long long a,long long x);int main(){int n,k;long long buf[10];long long answer;while (cin>>n){for (int time=0;time<n;time++){//initiate//inputfor (int i=0;i<3;i++)cin>>buf[i];cin>>k;if (buf[1]-buf[0]==buf[2]-buf[1])//arithmetic answer=(buf[0]%200907+( (buf[1]-buf[0])%200907 )*( (k-1)%200907 ))%200907;else{//geometric answer=(buf[0]%200907*(pow(buf[2]/buf[1],k-1)%200907))%200907;}cout<<answer<<endl;}}return true;}long long pow(long long a,long long x){long long pro=1;while(x>0){if (x%2==1){pro*=a;pro%=200907;}x/=2;a*=a;a%=200907;}return pro;}