HDU3483 A Very Simple Problem (矩阵快速幂)

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本文介绍了一种利用矩阵快速幂解决特定数列求和问题的方法。通过构造矩阵并运用快速幂运算，有效地解决了形如Σ(k^x*x^k) (1≤k≤n)的数列求和问题，并提供了完整的代码实现。

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A Very Simple Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1112 Accepted Submission(s): 549

Problem Description

This is a very simple problem. Given three integers N, x, and M, your task is to calculate out the following value:

Input

There are several test cases. For each case, there is a line with three integers N, x, and M, where 1 ≤ N, M ≤ 2*109, and 1 ≤ x ≤ 50.
The input ends up with three negative numbers, which should not be processed as a case.

Output

For each test case, print a line with an integer indicating the result.

Sample Input

100 1 10000 3 4 1000 -1 -1 -1

Sample Output

5050 444

Source

2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3483

题意：就是让求 segma(k^x*x^k)(1<=k<=n)的和。

思路：明显的要用矩阵快速幂求解，假设Sn = segma(k^x * x^k)(1<=k<=n),则Sn = Sn-1 + (n^x * x^n),令Xn-1 = {Sn-1,(n-1)^x........1}T,其中除了第一项外其余项都要*X^n，然后可以求出系数矩阵A

然后就可以对ori.a数组进行初始化操作，注意初始化的时候除了ori.a[0][0]外，其余都要*x，原因上面已经说过就不再啰嗦。当n=2时，得出的才是x1，所以Xn = A*Xn-1，求得的为Sn-1,Xn+1求出的才是Sn，所以就是Xn+1 = A^n * X1。

代码如下：

#include <bits/stdc++.h>
using namespace std;
#define ll long long

struct Matrix{
	ll a[55][55];
}ori,res;
ll n,m,x,c[55][55];

void Init(){
	memset(c,0,sizeof(c));
	c[0][0] = 1;
	for(int i = 1; i <= x; i ++){
		for(int j = 0; j <= i; j ++){
			if(j == 0 || j == i) c[i][j] = 1;
			else c[i][j] = c[i-1][j-1] + c[i-1][j];
		}
	}
	
	memset(ori.a,0,sizeof(ori.a));
	ori.a[0][0] = 1;
	for(int i = 1; i <= x+1; i ++){
		ori.a[0][i] = c[x][i-1] * x % m;
	}
	for(int i = 1; i <= x+1; i ++){
		for(int j = i,k = 0; j <= x+1; j ++,k ++){
			ori.a[i][j] = c[x-i+1][k] * x % m;
		}
	}
		
	memset(res.a,0,sizeof(res.a));
	for(int i = 0; i <= x+1; i ++)
		res.a[i][i] = 1;
}

Matrix multiply(Matrix b, Matrix d){
	Matrix temp;
	memset(temp.a,0,sizeof(temp.a));
	for(int i = 0; i <= x+1; i ++){
		for(int j = 0; j <= x+1; j ++){
			for(int k = 0; k <= x+1; k ++){
				temp.a[i][j] = (temp.a[i][j] + b.a[i][k] * d.a[k][j]) % m;
			}
		}
	}
	return temp;
}

void calcu(int n){
	while(n){
		if(n & 1) res = multiply(res,ori);
		n >>= 1;
		ori = multiply(ori,ori);
	}

	printf("%lld\n",(res.a[0][x+1] + m) % m);
}

int main(){
	while(~scanf("%lld%lld%lld",&n,&x,&m)){
		if(n < 0 && m < 0 && x < 0) break;
		Init();
		calcu(n);
	}
	return 0;
}