Hackonacci Matrix Rotations 观察题 ，更新了我的模板

https://www.hackerrank.com/contests/w27/challenges/hackonacci-matrix-rotations

一开始是没想到观察题的。只想到直接矩阵快速幂。

但是超时了，因为我的矩阵快速幂是应对稀疏矩阵的，

http://www.cnblogs.com/liuweimingcprogram/p/5942591.html

因为当时做这题的时候，以前的矩阵快速幂模板超时，所以就换了个，但是这题用稀疏矩阵的模板会超时，所以就把两个模板都收录算了。但是时间是1.58s，现在是pertest。所以后面的可能会超时，然而这题应该直接观察，其实也不算观察，算数学吧。鸽笼原理

前3项固定，最多1/2 * 1/2 * 1/2 项后，就会和前3项重复。

然后算一算，就发现了重复了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
 
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
int n;
const int maxn =  + ;
char str[maxn][maxn];
int ans[];
int f[] = {, , , , , , , };
void init() {
    for (int i = ; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {
            if (i ==  && j == ) continue;
            LL val = (1LL * i * j) * (1LL * i * j);
            val %= ;
            if (val == ) val = ;
            if (f[val] == ) {
                str[i][j] = 'Y';
            } else str[i][j] = 'X';
        }
//        cout << endl;
    }
    for (int i = ; i <= n; ++i) {
        for (int j = ; j <= i - ; ++j) {
            str[i][j] = str[j][i];
        }
    }
//    for (int i = 1; i <= n; ++i) {
//        for (int j = 1; j <= n; ++j) {
//            cout << str[i][j];
//        }
//        cout << endl;
//    }
    //
    int p1 = n, p2 = ;
    for (int i = ; i <= n; ++i) {
        p1 = n;
        p2 = i;
        for (int j = ; j <= n; ++j) {
            if (str[i][j] != str[p1--][p2]) {
                ans[]++;
            }
        }
        p1 = n - i + ;
        p2 = n;
        for (int j = ; j <= n; ++j) {
            if (str[i][j] != str[p1][p2--]) {
                ans[]++;
            }
        }
        p1 = n - i + ;
        for (int j = ; j <= n; ++j) {
            if (str[i][j] != str[p1][j]) {
                ans[]++;
            }
        }
    }
 
}
void work() {
    int q;
    scanf("%d%d", &n, &q);
    init();
    while (q--) {
        int x;
        scanf("%d", &x);
//        cin >> x;
        x %= ;
        x /= ;
//        cout << ans[x] << endl;
        printf("%d\n", ans[x]);
    }
}
 
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return ;
}

观察，0ms

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
 
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
int n;
const int maxn =  + ;
char str[maxn][maxn];
struct Matrix {
    LL a[][];
    int row;
    int col;
};
//struct Matrix matrix_mul(struct Matrix a, struct Matrix b, int MOD) {
//    struct Matrix c = {0};
//    c.row = a.row;
//    c.col = b.col;
//    for (int i = 1; i <= a.row; ++i) {
//        for (int k = 1; k <= a.col; ++k) {
//            if (a.a[i][k]) {
//                for (int j = 1; j <= b.col; ++j) {
//                    c.a[i][j] += a.a[i][k] * b.a[k][j];
////                    c.a[i][j] = (c.a[i][j] + MOD) % MOD;
//                }
//                c.a[i][j] %= MOD;
//            }
//        }
//    }
//    return c;
//}
struct Matrix matrix_mul  (struct Matrix a, struct Matrix b, int MOD) {
    struct Matrix c = {};
 
    c.row = a.row;
    c.col = b.col;
    for (int i = ; i <= a.row; i++) {
        for (int j = ; j <= b.col; j++) {
            for (int k = ; k <= b.row; k++) {
                c.a[i][j] += a.a[i][k] * b.a[k][j];
            }
            c.a[i][j] = (c.a[i][j] + MOD) % MOD;
        }
    }
    return c;
}
 
struct Matrix quick_matrix_pow(struct Matrix ans, struct Matrix base, LL n, int MOD) {
    while (n) {
        if (n & ) {
            ans = matrix_mul(ans, base, MOD);
        }
        n >>= ;
        base = matrix_mul(base, base, MOD);
    }
    return ans;
}
int ans[];
void init() {
    struct Matrix A;
    A.row = ;
    A.col = ;
    A.a[][] = ;
    A.a[][] = ;
    A.a[][] = ;
    struct Matrix B;
    B.col = B.row = ;
    B.a[][] = ;
    B.a[][] = ;
    B.a[][] = ;
    B.a[][] = ;
    B.a[][] = ;
    B.a[][] = ;
    B.a[][] = ;
    B.a[][] = ;
    B.a[][] = ;
    str[][] = 'Y';
//    cout << "1" << " ";
    for (int i = ; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {
            if (i ==  && j == ) continue;
            LL val = (1LL * i * j) * (1LL * i * j);
//            cout << val << " ";
            struct Matrix t = quick_matrix_pow(A, B, val - , );
//            cout << B.a[2][3] << " ";
            if (t.a[][] & ) {
                str[i][j] = 'Y';
            } else str[i][j] = 'X';
//            cout << t.a[1][3] << " ";
        }
//        cout << endl;
    }
    for (int i = ; i <= n; ++i) {
        for (int j = ; j <= i - ; ++j) {
            str[i][j] = str[j][i];
        }
    }
//    for (int i = 1; i <= n; ++i) {
//        for (int j = 1; j <= n; ++j) {
//            cout << str[i][j];
//        }
//        cout << endl;
//    }
    //
    int p1 = n, p2 = ;
    for (int i = ; i <= n; ++i) {
        p1 = n;
        p2 = i;
        for (int j = ; j <= n; ++j) {
            if (str[i][j] != str[p1--][p2]) {
                ans[]++;
            }
        }
        p1 = n - i + ;
        p2 = n;
        for (int j = ; j <= n; ++j) {
            if (str[i][j] != str[p1][p2--]) {
                ans[]++;
            }
        }
        p1 = n - i + ;
        for (int j = ; j <= n; ++j) {
            if (str[i][j] != str[p1][j]) {
                ans[]++;
            }
        }
    }
 
}
void work() {
    int q;
    cin >> n >> q;
    init();
    while (q--) {
        int x;
        cin >> x;
        x %= ;
        x /= ;
        cout << ans[x] << endl;
    }
}
 
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    IOS;
    work();
    return ;
}

直接矩阵快速幂，1580ms