HDU 2276 Kiki & Little Kiki 2

最新推荐文章于 2024-07-30 18:36:42 发布

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最新推荐文章于 2024-07-30 18:36:42 发布

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分类专栏：矩阵快速幂文章标签：矩阵快速幂

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Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2742 Accepted Submission(s): 1447

Problem Description

There are n lights in a circle numbered from 1 to n. The left of light 1 is light n, and the left of light k (1< k<= n) is the light k-1.At time of 0, some of them turn on, and others turn off.
Change the state of light i (if it's on, turn off it; if it is not on, turn on it) at t+1 second (t >= 0), if the left of light i is on !!! Given the initiation state, please find all lights’ state after M second. (2<= n <= 100, 1<= M<= 10^8)

Input

The input contains one or more data sets. The first line of each data set is an integer m indicate the time, the second line will be a string T, only contains '0' and '1' , and its length n will not exceed 100. It means all lights in the circle from 1 to n.
If the ith character of T is '1', it means the light i is on, otherwise the light is off.

Output

For each data set, output all lights' state at m seconds in one line. It only contains character '0' and '1.

Sample Input

Sample Output

Source

HDU 8th Programming Contest Site（1）

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lcy

题意：操作n次，每次操作将为1的灯的右边灯反转，求最后灯的状态

每次的结果取决于左边，可以简单列举几个，1 1 第二个灯的状态会变成0， 1 0 第二个灯的状态会变成1，0 0 地二个灯的状态还是0 ，从矩阵上考虑可以构造一个n*n的矩阵，对角线和对角线前一位为1，其余均为0，运算上如果每次%2效率会很低，只有0 1 两种情况就可以特殊处理了，可以把+和*的运算换位^和&的运算，因为矩阵每次都能从上一行经过变化得到，所以通过这一点能将复杂度从n^3降到n^

需要构造的矩阵为：

1 0 0 0 0 0 ...... 1 a[0]

1 1 0 0 0 0 ...... 0 a[1]

0 1 1 0 0 0 ...... 0 a[2]

0 0 0 1 1 0 ...... 0 X a[3]

................... ....

0 0 0 0 0 0 ..... 1 0 a[n-1]

#pragma comment(linker,"/STACK:1024000000,1024000000")
#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;
char g[110];
int n,a[110],ans[110][110],p[110][110],len;

void multi(int a[][110],int b[][110])
{
    int tmp[110][110];
    mem(tmp,0);
    for0(i,len)
        for0(j,len)
            tmp[0][i]^=a[0][j]&b[j][i];
    for(int i=1;i<len;i++)
        for0(j,len)
            a[i][j]=tmp[i][j]=tmp[i-1][(j+len-1)%len];//每一行的第n个可根据前一个递推得到
    for0(i,len)a[0][i]=tmp[0][i];
}

void init()//矩阵初始化
{
    mem(ans,0);
    for0(i,len)ans[i][i]=1;
    mem(p,0);
    for0(i,len)p[i][i]=p[i][(i+len-1)%len]=1;
}

void solve()//快速幂
{
    while(n)
    {
        if(n&1)multi(ans,p);
        n>>=1;
        multi(p,p);
    }
}

int main()
{
    while(~sf("%d",&n))
    {
        sf("%s",g);
        len=strlen(g);
        for0(i,len)a[i]=g[i]-'0';
        init();
        solve();
        for0(i,len)
        {
            int o=0;
            for0(j,len)
                o^=ans[i][j]&a[j];//计算每一个的情况
            pf(i==len-1?"%d\n":"%d",o);
        }
    }
    return 0;
}