【动态规划】Sicily1280 Permutation

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http://soj.sysu.edu.cn/show_problem.php?pid=1280

dp[i][j]表示已经填了i个数，剩余有j个数比填好的第i个数小。

转移：

dp[i + 1][k] += dp[i][j];

a[i] > a[i + 1] k: 0 to j - 1

a[i] < a[i + 1] k: j to n - i - 1

起始状态：dp[1][i] = 1 (i : 0 to n - 1)

最终结果：dp[n][0]

注意每次清空数组。

#include <cstdio>#include <iostream>#include <cstring>using namespace std;int n, p[105];long long f[105][105];int main(){    int a, b;    while (scanf("%d", &n) && n) {        memset(f, 0, sizeof(f));        scanf("%d", &a);        for (int i = 1; i < n; ++ i) {            scanf("%d", &b); p[i] = (a < b);            a = b;        }        for (int i = 0; i < n; ++ i) f[1][i] = 1;        for (int i = 1; i < n; ++ i) {            for (int j = 0; j <= n - i; ++ j) {                if (!p[i]) for (int k = 0; k <= j - 1; ++ k) {                    f[i + 1][k] += f[i][j];                } else for (int k = j; k <= n - i - 1; ++ k) {                    f[i + 1][k] += f[i][j];                }            }        }        printf("%lld\n", f[n][0]);    }    return 0;}

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