FZU Problem 1692 Key problem(迴圈矩陣)

畫船聽雨發表於2014-10-07
矩陣

迴圈矩陣，這裡有講解：http://wenku.baidu.com/link?url=zcJ-sxrj0QDqzz8xCnHTnB7gxjoNRyOZzS4_4ZA22c8Bs9inYn6vVkqTVr_w-riLa8oRnYA9SRcCZ9f4UciCUNGeNAG4dCGclYRPS18YLGa

推出第一層下面根據性質就可以得到。

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月賽-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;
}


相關文章