hdu5895Mathematician QSC+矩陣快速冪+除法取餘
Problem Description
QSC dream of becoming a mathematician, he believes that everything in this world has a mathematical law.
Through unremitting efforts, one day he finally found the QSC sequence, it is a very magical sequence, can be calculated by a series of calculations to predict the results of a course of a semester of a student.
This sequence is such like that, first of all,
QSC sequence published caused a sensation, after a number of students to find out the results of the prediction is very accurate, the shortcoming is the complex calculation. As clever as you are, can you write a program to predict the mark?
Input
First line is an integer T(1≤T≤1000).
The next T lines were given n, y, x, s, respectively.
n、x is 8 bits decimal integer, for example, 00001234.
y is 4 bits decimal integer, for example, 1234.
n、x、y are not negetive.
1≤s≤100000000
Output
For each test case the output is only one integer number ans in a line.
Sample Input
2
20160830 2016 12345678 666
20101010 2014 03030303 333
Sample Output
1
317
Source
2016 ACM/ICPC Asia Regional Shenyang Online
/*****************************/
….
累加
因為
所以
所以直接構造fn的矩陣快速冪就可以求得了。。。
然後就是兩個小結論要用到。。(orz,窩弱只知道其中一個啊,全場wa到死。。)
1.高次冪取膜
φ(p)為p的尤拉函式
2.除法取膜
所以小範圍爆一下,大點>φ(p)就用矩陣優化。。。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
//#define MOD 10000007
LL MOD;
struct Mat{
int n,m;
LL mat[9][9];
};
Mat operator *(Mat a,Mat b){
Mat c;
memset(c.mat,0,sizeof(c.mat));
c.n = a.n,c.m = b.m;
for(int i=1;i<=a.n;i++){
for(
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