B - The Staircases (dp)

标签：mount   sts   状态   printf   ++   write   contains   nta   ons   

One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps of different sizes in a strictly descending order. It is not allowed for staircase to have steps equal sizes. Every staircase consists of at least two steps and each step contains at least one brick. Picture gives examples of staircase for N=11 and N=5:

Your task is to write a program that reads from input numbers N and writes to output numbers Q - amount of different staircases that can be built from exactly N bricks.

Input

Numbers N, one on each line. You can assume N is between 3 and 500, both inclusive. A number 0 indicates the end of input.

<b< dd="">Output

Numbers Q, one on each line.

<b< dd="">Sample Input

3 
5 
0 

状态转移方程：dp[i][j]=dp[i-j][j-1]+dp[i][j-1];

注意开数组不要太大，可能会T

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
ll dp[1500][1500];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        break;
        memset(dp,0,sizeof(dp));
        for(int i=0;i<=n;i++) 
            dp[0][i]=1;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(i>=j)
                    dp[i][j]=dp[i-j][j-1]+dp[i][j-1];
                else
                    dp[i][j]=dp[i][j-1];
        printf("%lld\n",dp[n][n]-1);
    }
    return 0;
}

B - The Staircases (dp)

标签：mount   sts   状态   printf   ++   write   contains   nta   ons   

原文地址：https://www.cnblogs.com/Staceyacm/p/10822360.html