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
     
       #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
     
       #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
     
       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; }