FZU 1692 Key problem

 Problem 1692 Key problem

Accept: 208    Submit: 892
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

题意：每次按a[i]=(a[i]+L*a[(i+1)%n]+R*a[(i-1+n)%n])%M(原题打错了）,操作m次后n个数的结果

矩阵很好想，自己做的时候想对了，竟然T了，自己还是太水了

1 L 0 0 0 0 R        a[0]

R 1 L 0 0 0 0        a[1]

0 R 1 L 0 0 0        a[2]

.  .  .  .  . .  .         a[n-1](依次类推）

写矩阵乘法的时候如果直接3重循环的话，假设数据范围是100X100的矩阵，就要花1e6的时间，肯定T了，通过观察可以发现，矩阵第一行和第二行变化规律一样，可以直接有第一个转换过来

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<stack>#include<queue>#include<deque>#include<set>#include<map>#include<cmath>#include<vector>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> PII;#define pi acos(-1.0)#define eps 1e-10#define pf printf#define sf scanf#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r#define e tree[rt]#define _s second#define _f first#define all(x) (x).begin,(x).end#define mem(i,a) memset(i,a,sizeof i)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define mi ((l+r)>>1)#define sqr(x) ((x)*(x))const int inf=0x3f3f3f3f;int t,M,n,m,L,R;ll ans[110][110],a[110],p[110][110];void multi(ll a[][110],ll b[][110])//矩阵乘法{    int tmp[110][110];    mem(tmp,0);    for0(i,n)//不能套三层循环，不然T了        for0(j,n)            tmp[0][i]=(tmp[0][i]+a[0][j]*b[j][i]%M)%M;    for(int i=1;i<n;i++)        for0(j,n)            tmp[i][j]=tmp[i-1][(j-1+n)%n];    for0(i,n)        for0(j,n)            a[i][j]=tmp[i][j];}void init(){    mem(ans,0);    mem(p,0);    for0(i,n)        ans[i][i]=1;    for0(i,n)        p[i][i]=1,p[i][(i+1)%n]=L,p[i][(i+n-1)%n]=R;}int main(){    sf("%d",&t);    while(t--)    {        sf("%d%d%d%d%d",&n,&m,&L,&R,&M);        init();//矩阵初始化        for0(i,n)            sf("%d",&a[i]);        while(m)//矩阵快速幂        {            if(m&1)multi(ans,p);            m>>=1;            multi(p,p);        }        for0(i,n)        {            int o=0;            for0(j,n)                o=(o+ans[i][j]*a[j]%M)%M;            pf(i!=n-1?"%d ":"%d\n",o);        }    }    return 0;}