FZU Problem 1692 Key problem(循环矩阵)

循环矩阵，这里有解说：http://wenku.baidu.com/link?

url=zcJ-sxrj0QDqzz8xCnHTnB7gxjoNRyOZzS4_4ZA22c8Bs9inYn6vVkqTVr_w-riLa8oRnYA9SRcCZ9f4UciCUNGeNAG4dCGclYRPS18YLGa

推出第一层以下依据性质就能够得到。

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;
}

时间： 2024-10-29 10:45:47