zoj1163 The Staircases （简单dp）

最新推荐文章于 2019-03-08 15:07:40 发布

a451907

最新推荐文章于 2019-03-08 15:07:40 发布

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本文链接：https://blog.youkuaiyun.com/a451907/article/details/8614674

本文介绍了一种使用动态规划算法解决砖块构建不同楼梯组合的问题。孩子利用N块砖搭建楼梯，楼梯至少包含两个台阶且每个台阶至少一块砖，台阶大小必须递减且不相等。文章提供了一个C++实现的示例程序，该程序能够计算给定数量的砖块可以搭建的不同楼梯数量。

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The Staircases

Time Limit: 2 Seconds Memory Limit: 65536 KB

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.

Output

Numbers Q, one on each line.

Sample Input

3
5
0

Sample Output

1
2

简单dp题，用整形会溢出。

#include<stdio.h>
#include<string.h>
long long a[505];
int main()
{
    memset(a,0,sizeof(a));
    long n;
    a[0]=1;
    for(long k=0;k<=500;k++)
        for(long i=500;i>0;i--)
            if(i-k>=0)
                a[i]+=a[i-k];
    while(scanf("%ld",&n)&&n)
        printf("%lld\n",a[n]-1);
    return 0;
}