hdu 5015(矩阵快速幂)

来源：互联网发布：windows打开81端口编辑：程序博客网时间：2024/05/19 19:43

题意：有一个矩阵横排第一行初始是0,233,2333,23333...，然后给出计算公式a[i][j] = a[i - 1][j] + a[i][j - 1]，然后给出了n和m和第一列的各值要求计算a[n][m]的值。

题解：n <= 10但m <= 10^9，空间和时间花费都大，已知第一列a0，a1，a2......an，那么第二列每个数字就是b1 = a1 + 233，b2 = b1 + a2 = a1 + a2 + 233... 找到特点后构造矩阵然后用矩阵快速幂求解。

#include <stdio.h>#include <string.h>#define ll long longconst int N = 15;const int MOD = 10000007;int n, m;struct Mat {    ll a[N][N];    Mat() {        memset(a, 0, sizeof(a));    }    Mat operator * (Mat b) {        Mat c;        for (int i = 0; i < n; i++)            for (int j = 0; j < n; j++)                for (int k = 0; k < n; k++)                    c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j]) % MOD;        return c;    }};Mat f(Mat b, int m) {    Mat c;    for (int i = 0; i < n; i++)        c.a[i][i] = 1;    while (m) {        if (m & 1)            c = c * b;        b = b * b;        m >>= 1;    }    return c;}int main() {    while (scanf("%d%d", &n, &m) == 2) {        Mat A;        A.a[0][0] = 1;        A.a[1][0] = 1;        A.a[1][1] = 10;        n += 2;        for (int i = 2; i < n; i++)            for (int j = 1; j <= i; j++)                A.a[i][j] = 1;        A = f(A, m);        ll temp[N], res = 0;        temp[0] = 3;        temp[1] = 233;        for (int i = 2; i < n; i++)            scanf("%lld", &temp[i]);        for (int i = 0; i < n; i++)            res = (res + temp[i] * A.a[n - 1][i] % MOD) % MOD;        printf("%lld\n", res);    }    return 0;}

0 0