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;
}