矩阵乘法 pku 3233 Matrix Power Series hdu 1757 A Simple Math Problem fzu 1683 纪念SlingShot hdu 3306 Another kind of Fibonacc

分析：矩阵相乘O(n^3), 有k次，则复杂度为n^3*k。

使用矩阵技巧，构造：
    B=  |  A  A  |
          |  O  I   |
则B的乘方的结果其右上会是S，其他三个不变。此时化成了矩阵乘方问题，此时可以使用反复平方法，这样复杂度为(2n)^3*logk

#include<stdio.h>
#define nnum 61
int res[nnum][nnum], te[nnum][nnum], n, m;
void mulmatrix(int a[nnum][nnum], int b[nnum][nnum], int mark) {
	int i, j, k, temp[nnum][nnum];
	for (i = 0; i < n; i++) {
		for (j = 0; j < n; j++) {
			temp[i][j] = 0;
			for (k = 0; k < n; k++) {
				temp[i][j] += a[i][k] * b[k][j];
				temp[i][j] %= m;
			}
		}
	}
	if (mark) {
		for (i = 0; i < n; i++) {
			for (j = 0; j < n; j++) {
				te[i][j] = temp[i][j];
			}
		}
	} else {
		for (i = 0; i < n; i++) {
			for (j = 0; j < n; j++) {
				res[i][j] = temp[i][j];
			}
		}
	}
}
void cal(int times) {
	while (times) {
		if (times & 1) {
			mulmatrix(te, res, 1);
		}
		mulmatrix(res, res, 0);
		times >>= 1;
	}
}
int main() {
#ifndef ONLINE_JUDGE
	freopen("t.txt", "r", stdin);
#endif
	int k, i, j, re;
	while (scanf("%d %d %d", &n, &k, &m) != EOF) {
		for (i = 0; i < n; i++) {
			for (j = 0; j < n; j++) {
				scanf("%d", &res[i][j]);
				res[i][j + n] = res[i][j];
			}
		}
		for (i = n; i < n + n; i++) {
			for (j = n; j < n; j++) {
				res[i][j] = 0;
			}
			res[i][i] = 1;
		}
		re = n;
		n = n + n;
		for (i = 0; i < n; i++) {
			for (j = 0; j < n; j++) {
				te[i][j] = 0;
			}
			te[i][i] = 1;
		}
		cal(k);
		for (i = 0; i < re; i++) {
			for (j = re; j < n; j++) {
				printf("%d ", te[i][j]);
			}
			printf("\n");
		}
	}
	return 0;
}

/*
 * hdu1757.c
 *
 *  Created on: 2011-9-12
 *      Author: bjfuwangzhu
 */
//
//#include<stdio.h>
//#define nmax 10
//typedef struct matrix {
//    int num[nmax][nmax];
//} matrix;
//int m;
//matrix res, temp;
//matrix matrix_mul(matrix numa, matrix numb) {
//    matrix numc;
//    int i, j, k;
//    for (i = 0; i < nmax; i++) {
//        for (j = 0; j < nmax; j++) {
//            numc.num[i][j] = 0;
//            for (k = 0; k < nmax; k++) {
//                numc.num[i][j] += numa.num[i][k] * numb.num[k][j];
//                numc.num[i][j] %= m;
//            }
//        }
//    }
//    return numc;
//}
//void solve(int k) {
//    while (k) {
//        if (k & 1) {
//            res = matrix_mul(res, temp);
//        }
//        temp = matrix_mul(temp, temp);
//        k >>= 1;
//    }
//}
//int main() {
//#ifndef ONLINE_JUDGE
//    freopen("data.in", "r", stdin);
//#endif
//    int k, i, j, ans;
//    while (scanf("%d %d", &k, &m) != EOF) {
//        for (i = 0; i < nmax; i++) {
//            for (j = 0; j < nmax; j++) {
//                temp.num[i][j] = 0;
//                res.num[i][j] = 0;
//            }
//            res.num[i][i] = 1;
//        }
//        for (i = 0; i < nmax; i++) {
//            scanf("%d", &temp.num[0][i]);
//        }
//        for (i = 0; i < nmax - 1; i++) {
//            temp.num[i + 1][i] = 1;
//        }
//        if (k < nmax) {
//            printf("%d\n", k % m);
//            continue;
//        }
//        solve(k - 9);
//        for (i = 0, ans = 0; i < nmax; i++) {
//            ans += res.num[0][i] * (nmax - 1 - i);
//            ans %= m;
//        }
//        printf("%d\n", ans);
//    }
//    return 0;
//}
#include<stdio.h>
#define nmax 10
int matrix_res[nmax][nmax], matrix_temp[nmax][nmax];
void matrix_mul(int a[nmax][nmax], int b[nmax][nmax], int m, int mark) {
    int temp[nmax][nmax];
    int i, j, k;
    for (i = 0; i < nmax; i++) {
        for (j = 0; j < nmax; j++) {
            temp[i][j] = 0;
            for (k = 0; k < nmax; k++) {
                temp[i][j] += a[i][k] * b[k][j];
                temp[i][j] %= m;
            }
        }
    }
    if (mark) {
        for (i = 0; i < nmax; i++) {
            for (j = 0; j < nmax; j++) {
                matrix_res[i][j] = temp[i][j];
            }
        }
    } else {
        for (i = 0; i < nmax; i++) {
            for (j = 0; j < nmax; j++) {
                matrix_temp[i][j] = temp[i][j];
            }
        }
    }
}
void solve(int k, int m) {
    if (k < 0) {
        return;
    }
    while (k) {
        if (k & 1) {
            matrix_mul(matrix_res, matrix_temp, m, 1);
        }
        matrix_mul(matrix_temp, matrix_temp, m, 0);
        k >>= 1;
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int k, m, i, j, res;
    while (scanf("%d %d", &k, &m) != EOF) {
        for (i = 0; i < nmax; i++) {
            for (j = 0; j < nmax; j++) {
                matrix_temp[i][j] = 0;
                matrix_res[i][j] = 0;
            }
            matrix_res[i][i] = 1;
        }
        for (i = 0; i < nmax; i++) {
            scanf("%d", &matrix_temp[0][i]);
        }
        for (i = 0; i < nmax - 1; i++) {
            matrix_temp[i + 1][i] = 1;
        }
        if (k < nmax) {
            printf("%d\n", k % m);
            continue;
        }
        solve(k - 9, m);
        for (i = 0, res = 0; i < nmax; i++) {
            res = res + matrix_res[0][i] * (nmax - i - 1);
            res %= m;
        }
        printf("%d\n", res);
    }
    return 0;
}

/*
 * fzu1683.c
 *
 *  Created on: 2011-9-12
 *      Author: bjfuwangzhu
 */

#include<stdio.h>
#define nnum 2009
#define nmax 4
typedef struct matrix {
    int num[nmax][nmax];
} matrix;
matrix res, temp;
int sum[nmax];
matrix matrix_mul(matrix numa, matrix numb) {
    matrix numc;
    int i, j, k;
    for (i = 0; i < nmax; i++) {
        for (j = 0; j < nmax; j++) {
            numc.num[i][j] = 0;
            for (k = 0; k < nmax; k++) {
                numc.num[i][j] += numa.num[i][k] * numb.num[k][j];
                numc.num[i][j] %= nnum;
            }
        }
    }
    return numc;
}
void solve(int k) {
    while (k) {
        if (k & 1) {
            res = matrix_mul(res, temp);
        }
        temp = matrix_mul(temp, temp);
        k >>= 1;
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int m, n, i, j, k, ant;
    while (scanf("%d", &m) != EOF) {
        for (i = 1; i <= m; i++) {
            scanf("%d", &n);
            for (j = 0; j < nmax; j++) {
                for (k = 0; k < nmax; k++) {
                    temp.num[j][k] = 0, res.num[j][k] = 0;
                }
                res.num[j][j] = 1;
            }
            temp.num[0][0] = 1, temp.num[0][1] = 3, temp.num[0][2] = 2, temp.num[0][3] =
                    7;
            temp.num[1][1] = 3, temp.num[1][2] = 2, temp.num[1][3] = 7;
            temp.num[2][1] = 1, temp.num[3][2] = 1;
            sum[3] = 1, sum[2] = 3, sum[1] = 5, sum[0] = 9;
            printf("Case %d: ", i);
            if (n < nmax - 1) {
                for (j = 0, ant = 0; j <= n; j++) {
                    ant += sum[nmax - 1 - j];
                    ant %= nnum;
                }
                printf("%d\n", ant);
                continue;
            }
            solve(n - 2);
            for (j = 0, ant = 0; j < nmax; j++) {
                ant += res.num[0][j] * sum[j];
                ant %= nnum;
            }
            printf("%d\n", ant);
        }
    }
    return 0;
}

/*
 C语言版
 */
#include<stdio.h>
#define nmax 4
#define nnum 10007
typedef struct matrix {
    int num[nmax][nmax];
} matrix;
matrix matrix_mul(matrix numa, matrix numb) {
    matrix numc;
    int i, j, k;
    for (i = 0; i < nmax; i++) {
        for (j = 0; j < nmax; j++) {
            numc.num[i][j] = 0;
            for (k = 0; k < nmax; k++) {
                numc.num[i][j] = (numc.num[i][j]
                        + numa.num[i][k] * numb.num[k][j]) % nnum;
            }
        }
    }
    return numc;
}
matrix matrix_exp(matrix numc, matrix numa, int x) {
    for (; x; x >>= 1) {
        if (x & 1) {
            numc = matrix_mul(numc, numa);
        }
        numa = matrix_mul(numa, numa);
    }
    return numc;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int i, j, N, X, Y, sum[nmax] = { 2, 1, 1, 1 }, ant;
    matrix res, temp;
    while (scanf("%d %d %d", &N, &X, &Y) != EOF) {
        if (N == 0) {
            printf("1\n");
            continue;
        }
        if (N == 1) {
            printf("2\n");
            continue;
        }
        for (i = 0; i < nmax; i++) {
            for (j = 0; j < nmax; j++) {
                res.num[i][j] = 0, temp.num[i][j] = 0;
            }
            res.num[i][i] = 1;
        }
        X %= nnum,Y %= nnum;
        temp.num[0][0] = 1, temp.num[0][1] = X * X % nnum, temp.num[0][2] = Y
                * Y % nnum, temp.num[0][3] = 2 * X * Y % nnum;
        temp.num[1][1] = X * X % nnum, temp.num[1][2] = Y * Y % nnum, temp.num[1][3] =
                2 * X * Y % nnum;
        temp.num[2][1] = 1, temp.num[3][1] = X, temp.num[3][3] = Y;
        res = matrix_exp(res, temp, (N - 1));
        for (i = 0, ant = 0; i < nmax; i++) {
            ant += res.num[0][i] * sum[i];
            ant %= nnum;
        }
        printf("%d\n", ant);
    }
    return 0;
}

/*
 * hdu3306.cpp
 *
 *  Created on: 2011-9-13
 *      Author: bjfuwangzhu
 */
/*
 C++语言版 重载运算符
 */
#include<stdio.h>
#define nmax 4
#define nnum 10007
struct matrix {
    int num[nmax][nmax];
    matrix() {
        for (int i = 0; i < nmax; i++) {
            for (int j = 0; j < nmax; j++) {
                num[i][j] = 0;
            }
        }
    }
};
matrix operator*(matrix numa, matrix numb) {
    matrix numc;
    int i, j, k;
    for (i = 0; i < nmax; i++) {
        for (j = 0; j < nmax; j++) {
            for (k = 0; k < nmax; k++) {
                numc.num[i][j] = (numc.num[i][j]
                        + numa.num[i][k] * numb.num[k][j]) % nnum;
            }
        }
    }
    return numc;
}
matrix operator^(matrix numa, int x) {
    matrix numc;
    int i;
    for (i = 0; i < nmax; i++) {
        numc.num[i][i] = 1;
    }
    for (; x; x >>= 1) {
        if (x & 1) {
            numc = numc * numa;
        }
        numa = numa * numa;
    }
    return numc;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int i, N;
    int X, Y, sum[nmax] = { 2, 1, 1, 1 }, ant;
    matrix res, temp;
    while (scanf("%d %d %d", &N, &X, &Y) != EOF) {
        if (N == 0) {
            printf("1\n");
            continue;
        }
        if (N == 1) {
            printf("2\n");
            continue;
        }
        X %= nnum,Y %= nnum;
        temp.num[0][0] = 1, temp.num[0][1] = X * X % nnum, temp.num[0][2] = Y
                * Y % nnum, temp.num[0][3] = 2 * X * Y % nnum;
        temp.num[1][1] = X * X % nnum, temp.num[1][2] = Y * Y % nnum, temp.num[1][3] =
                2 * X * Y % nnum;
        temp.num[2][1] = 1, temp.num[3][1] = X, temp.num[3][3] = Y;
        res = temp ^ (N - 1);
        for (i = 0, ant = 0; i < nmax; i++) {
            ant += res.num[0][i] * sum[i];
            ant %= nnum;
        }
        printf("%d\n", ant);
    }
    return 0;
}