hdu 1005 Number Sequence zoj 1105

使用矩阵的方法解决zoj2105

结果出现了一下两种情况：

仅仅是改了一下求第n项，变为先求第n-1,n-2项，再求第n项！结果竟是不一样~~路过的大牛指导一下，谢谢！！

Wrong的代码：

/*
 * zoj2105.c
 *
 *  Created on: 2011-9-20
 *      Author: bjfuwangzhu
 */

#include<stdio.h>
#define nmax 2
#define nnum 7
typedef struct matrix {
    int num[nmax][nmax];
} matrix;
matrix mul_matrix(matrix a, matrix b) {
    matrix c;
    int i, j, k;
    for (i = 0; i < nmax; i++) {
        for (j = 0; j < nmax; j++) {
            c.num[i][j] = 0;
            for (k = 0; k < nmax; k++) {
                c.num[i][j] += a.num[i][k] * b.num[k][j] % nnum;
                c.num[i][j] %= nnum;
            }
        }
    }
    return c;
}
void solve(int a, int b, int n) {
    matrix temp, res;
    res.num[0][0] = 1, res.num[1][1] = 1, res.num[1][0] = res.num[0][1] = 0;
    temp.num[0][0] = a % nnum, temp.num[0][1] = b % nnum, temp.num[1][0] = 1, temp.num[1][1] =
            0;
    while (n) {
        if (n & 1) {
            res = mul_matrix(res, temp);
        }
        temp = mul_matrix(temp, temp);
        n >>= 1;
    }
    printf("%d\n", res.num[0][0]);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int a, b, n;
    while (~scanf("%d %d %d", &a, &b, &n)) {
        if (a == 0 && b == 0 && n == 0) {
            break;
        }
        if (n == 1 || n == 2) {
            puts("1");
            continue;
        }
        solve(a, b, n - 1);
    }
    return 0;
}

Accept的代码：

/*
 * zoj2105.c
 *
 *  Created on: 2011-9-20
 *      Author: bjfuwangzhu
 */

#include<stdio.h>
#define nmax 2
#define nnum 7
typedef struct matrix {
    int num[nmax][nmax];
} matrix;
matrix mul_matrix(matrix a, matrix b) {
    matrix c;
    int i, j, k;
    for (i = 0; i < nmax; i++) {
        for (j = 0; j < nmax; j++) {
            c.num[i][j] = 0;
            for (k = 0; k < nmax; k++) {
                c.num[i][j] += a.num[i][k] * b.num[k][j] % nnum;
                c.num[i][j] %= nnum;
            }
        }
    }
    return c;
}
void solve(int a, int b, int n) {
    matrix temp, res;
    res.num[0][0] = 1, res.num[1][1] = 1, res.num[1][0] = res.num[0][1] = 0;
    temp.num[0][0] = a % nnum, temp.num[0][1] = b % nnum, temp.num[1][0] = 1, temp.num[1][1] =
            0;
    while (n) {
        if (n & 1) {
            res = mul_matrix(res, temp);
        }
        temp = mul_matrix(temp, temp);
        n >>= 1;
    }
    printf("%d\n", (res.num[0][0] + res.num[0][1]) % nnum);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int a, b, n;
    while (~scanf("%d %d %d", &a, &b, &n)) {
        if (a == 0 && b == 0 && n == 0) {
            break;
        }
        if (n == 1 || n == 2) {
            puts("1");
            continue;
        }
        solve(a, b, n - 2);
    }
    return 0;
}

另一种写法，找规律：

#include<iostream>
using namespace std;
int main() {
    int A, B, n, i, a[2000];
    while (cin >> A >> B >> n) {
        if (A == 0 && B == 0 && n == 0)
            break;
        if (n == 1 || n == 2) {
            cout << "1" << endl;
            continue;
        }
        a[1] = 1;
        a[2] = 1;
        A %= 7;
        B %= 7;
        for (i = 3; i < 2000; i++) {
            a[i] = (A * a[i - 1] + B * a[i - 2]) % 7;
            if (a[i - 1] == 1 && a[i] == 1)
                break;
        }
        i -= 2;
        n %= i;
        a[0] = a[i];
        cout << a[n] << endl;
    }
    return 0;
}