HDU 1292 "下沙野骆驼"ACM夏令营（DP）

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1292

思路：dp[i][j]表示前j个人分成i队，dp[i][j] = dp[i-1][j-1] + dp[i][j-1] * i表示第j个人单独成一队，或者在前i队中选择一个插入

AC代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long LL;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;LL dp[30][30];int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(dp,0,sizeof(dp));        for(int i=1; i<=n; i++)            dp[1][i] = dp[i][i] = 1;        LL sum = 1;//加上dp[1][n]的值        for(int i=2; i<=n; i++)        {            for(int j=i+1; j<=n; j++)            {                dp[i][j] = dp[i-1][j-1] + dp[i][j-1] * i;            }            sum += dp[i][n];        }        printf("%I64d\n",sum);    }    return 0;}