ural 1017. Staircases(dp)

http://acm.timus.ru/problem.aspx?space=1&num=1017

题意：有n块砖，要求按照严格递增的个数摆放成楼梯，求楼梯的摆放种类数。

思路：状态转移方程：dp[i][j]=sum(dp[i-j][k]), 0 < k < j; i 表示砖的总数，j表示最高的那层的个数。

#include <stdio.h>
#include <string.h>
#define LL long long
const int N=502;
LL dp

;
int main()
{
int n;
while(~scanf("%d",&n))
{
memset(dp,0,sizeof(dp));
for (int i = 1; i <= n; i++)
dp[i][i] = 1;
dp[3][2] = 1;
dp[4][3] = 1;
for (int i = 5; i <= n; i++)
{
for (int j = 1; j < i; j++)
{
for (int k = 1; k < j; k++)
dp[i][j]+=dp[i-j][k];
}
}
LL ans = 0;
for (int i = 1; i < n; i++)
ans+=dp
[i];
printf("%lld\n",ans);
}
return 0;
}

View Code