FZU Problem 1692 Key problem(循环矩阵)

最新推荐文章于 2019-01-08 08:27:05 发布

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本文探讨了循环矩阵的概念及其应用，通过实例演示如何利用循环矩阵解决特定问题，并提供了相应的算法实现。其中包括了问题描述、输入输出规范、示例输入输出及关键代码片段，旨在帮助读者理解和解决类似问题。

循环矩阵，这里有讲解：http://wenku.baidu.com/link?url=zcJ-sxrj0QDqzz8xCnHTnB7gxjoNRyOZzS4_4ZA22c8Bs9inYn6vVkqTVr_w-riLa8oRnYA9SRcCZ9f4UciCUNGeNAG4dCGclYRPS18YLGa

推出第一层下面根据性质就可以得到。

Problem 1692 Key problem

Accept: 144 Submit: 663
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

Whenever rxw meets Coral, he requires her to give him the laboratory key. Coral does not want to give him the key, so Coral ask him one question. if rxw can solve the problem, she will give him the key, otherwise do not give him. rxw turns to you for help now,can you help him?

N children form a circle, numbered 0,1,2, ... ..., n-1,with Clockwise. Initially the ith child has Ai apples. Each round game, the ith child will obtain ( L*A_(i+n-1)%n+R*A_(i+1)%n ) apples. After m rounds game, Coral would like to know the number of apples each child has. Because the final figure may be very large, so output the number model M.

Input

The first line of input is an integer T representing the number of test cases to follow. Each case consists of two lines of input: the first line contains five integers n,m,L,R and M . the second line contains n integers A0, A1, A2 ... An-1. (0 <= Ai <= 1000,3 <= n <= 100,0 <= L, R <= 1000,1 <= M <= 10 ^ 6,0 <=m < = 10 ^ 9). After m rounds game, output the number model M of apples each child has.

Output

Each case separated by a space. See sample.

Sample Input

1 3 2 3 4 10000 1 2 3

Sample Output

120 133 131

Source

FOJ月赛-2009年3月--- Coral

#include <set>
#include <map>
#include <queue>
#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

#define eps 1e-8
#define pi acos(-1.0)

#define LL __int64

using namespace std;


const int maxn = 110;

LL a[maxn], b[maxn], f[maxn];

LL mod, n;

void mul(LL a[], LL b[])
{
    LL c[maxn];
    memset(c, 0, sizeof(c));
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++) c[i] = (a[j]*b[(i-j+n)%n]+c[i])%mod;
    memcpy(a, c, sizeof(c));
}

void pow_mod(LL a[], LL b)
{
    LL c[maxn];
    memset(c, 0, sizeof(c));
    c[0] = 1LL;
    while(b)
    {
        if(b&1) mul(c, a);
        mul(a, a);
        b >>= 1;
    }
    memcpy(a, c, sizeof(c));
}

int main()
{
    int T;
    cin>>T;
    LL m, l, r;
    while(T--)
    {
        cin>>n>>m>>l>>r>>mod;
        for(int i = 0;i < n;i++)  cin>>a[i];
        memset(f, 0, sizeof(f));
        f[0] = 1; f[1] = r; f[n-1]=l;
        pow_mod(f, m);
        LL ans[maxn];
        for(int i = 0; i < n; i++)
        {
            ans[i] = 0;
            for(int j = 0;j < n;j++)
                ans[i] = (ans[i]+a[j]*f[(i-j+n)%n])%mod;
        }
        cout<<ans[0];
        for(int i = 1; i < n; i++) cout<<" "<<ans[i];
        cout<<endl;
    }
    return 0;
}