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;
}