HDU2276 Kiki & Little Kiki 2 矩阵快速幂

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3078    Accepted Submission(s): 1642

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

1 0101111 10 100000001

 

 

Sample Output

1111000 001000010

 

 

Source

HDU 8th Programming Contest Site（1）

 

 

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题意：给你n个灯泡，每个灯泡根据其左边的灯泡的开关（1代表开，0代表关）来决定自身是否要发生改变，开就发生改变，关则不变，1号灯泡的左边是n号灯泡，问经过m秒后n个灯泡是开还是关

 

分析：因为m可以很大，所以这里得用到矩阵快速幂做题。

这是另外一种形式的快速幂。首先当然是先找递推式：f(n) = ( f(n-1) + f(n) ) % 2，得到1代表开，0代表关

然后我们就可以写出矩阵式，不难得到：

| f(1)  f(2)  ...  f(n) |　　 　　| f(1)  f(2)  ...  f(n) |

| 0       0    ...    0  |　　x　　| 0       0    ...    0  |

| ...     ...    ...   ...  |　　 　　| ...     ...    ...   ...  |

| 0       0    ...    0  |　　 　　| 0       0    ...    0  |

然后根据矩阵式套矩阵快速幂模板就行

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 110;  //注意空间的大小，开多了会炸程序，尤其在程序中有多个矩阵的时候，最好开到刚刚符合题目要求
const int mod = 2;
typedef long long ll;
struct matrix {
    ll a[maxn][maxn];
};
matrix base, ans;
ll n;
string s;
matrix multip( matrix x, matrix y ) {
    matrix tmp;
    for( ll i = 0; i < s.length(); i ++ ) {
        for( ll j = 0; j < s.length(); j ++ ) {
            tmp.a[i][j] = 0;
            for( ll k = 0; k < s.length(); k ++ ) {
                tmp.a[i][j] = ( tmp.a[i][j] + x.a[i][k] * y.a[k][j] + mod ) % mod;
            }
        }
    }
    return tmp;
}
void f( ll x ) {
    while( x ) {
        if( x&1 ) {
            ans = multip( ans, base );
        }
        base = multip( base, base );
        x /= 2;
    }
}
int main() {
    while( cin >> n >> s ) {
        memset( ans.a, 0, sizeof(ans.a) );
        memset( base.a, 0, sizeof(base.a) );
        for( ll i = 0; i < s.length(); i ++ ) {
            ans.a[0][i] = s[i]-'0';
        }
        for( ll i = 0; i < s.length(); i ++ ) {
            if( i != s.length()-1 ) {
                base.a[i][i] = base.a[i][i+1] = 1;
            } else {
                base.a[i][i] = base.a[i][0] = 1;
            }
        }
        f(n);
        for( ll i = 0; i < s.length(); i ++ ) {
            cout << ans.a[0][i];
        }
        cout << endl;
    }
    return 0;
}