hdu 2276矩阵快速幂

根据题意构造变换矩阵即可。有很多构造矩阵的方法。例如，假设原串长度为7，则我构造的矩阵M如下：

1 1 0 0 0 0 0

0 1 1 0 0 0 0

0 0 1 1 0 0 0

0 0 0 1 1 0 0

0 0 0 0 1 1 0

0 0 0 0 0 1 1

1 0 0 0 0 0 1

每次用M右乘原始矩阵就相当于一次变换，用快速幂就可以过了。不过这道题测试数据有点坑爹，我直接用的矩阵模板，超时，手动把矩阵乘法部分改成位运算形式就过了。。。估计之前超时也就超那么一丁点吧，估计加个输入外挂也能过。。。

/*

 * hdu2276/win.cpp

 * Created on: 2012-11-3

 * Author    : ben

 */

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <queue>

#include <set>

#include <map>

#include <stack>

#include <string>

#include <vector>

#include <deque>

#include <list>

#include <functional>

#include <numeric>

#include <cctype>

using namespace std;

const int MAX_ORDER = 105;

typedef bool typec;

typedef struct MyMatrix {

    int order;

    typec num[MAX_ORDER][MAX_ORDER];

    MyMatrix(int ord) {

        order = ord;

    }

    void init() {

        for (int i = 0; i < order; i++) {

            for (int j = 0; j < order; j++) {

                num[i][j] = 0;

            }

        }

    }

} MyMatrix;

MyMatrix operator*(MyMatrix ma, MyMatrix mb) {

    int ord = ma.order;

    MyMatrix numc(ord);

    numc.init();

    int i, j, k;

    for (i = 0; i < ord; i++) {

        for (j = 0; j < ord; j++) {

            for (k = 0; k < ord; k++) {

                numc.num[i][j] = numc.num[i][j] xor (ma.num[i][k] and mb.num[k][j]);

            }

        }

    }

    return numc;

}

MyMatrix mpow(MyMatrix ma, int x) {

    int ord = ma.order;

    MyMatrix numc(ord);

    numc.init();

    for (int i = 0; i < ord; i++) {

        numc.num[i][i] = 1;

    }

    for (; x; x >>= 1) {

        if (x & 1) {

            numc = numc * ma;

        }

        ma = ma * ma;

    }

    return numc;

}



MyMatrix getTraMatix(int n) {

    MyMatrix ret(n);

    ret.init();

    for(int j = 0; j < n; j++) {

        int i = (j - 1 + n) % n;

        ret.num[j][j] = 1;

        ret.num[i][j] = 1;

    }

    return ret;

}



int main() {

#ifndef ONLINE_JUDGE

    freopen("data.in", "r", stdin);

#endif

    int m;

    char str[200];

    while(scanf("%d", &m) == 1) {

        getchar();

        gets(str);

        int len = strlen(str);

        MyMatrix ori(len);

        ori.init();

        for(int i = 0; i < len; i++) {

            ori.num[0][i] = str[i] - '0';

        }

        MyMatrix tra = mpow(getTraMatix(len), m);

        MyMatrix result = ori * tra;

        for(int i = 0; i < len; i++) {

            printf("%d", result.num[0][i]);

        }

        putchar('\n');

    }

    return 0;

}