HDU 1028 Ignatius and the Princess III

时间：2014-12-08 22:43:08 阅读：325 评论：0 收藏：0 [点我收藏+]

Description

"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:
  N=a[1]+a[2]+a[3]+...+a[m];
  a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
  4 = 4;
  4 = 3 + 1;
  4 = 2 + 2;
  4 = 2 + 1 + 1;
  4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

Input

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

Sample Input

4 10 20

Sample Output

5 42 627

代码是zxp那搬来的完全不明觉厉= =

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
__int64 a[150];
int main()
{
    int i,j;
    a[0]=1;//a[1]=1;a[2]=2;a[3]=3;a[4]=5;a[5]=7;
    for(i=1;i<=120;i++)
    {
        for(j=0;i+j<=120;j++)
        {
            a[i+j]+=a[j];
        }
    }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
       printf("%I64d\n",a[n]);
    }
    return 0;
}

View Code

HDU 1028 Ignatius and the Princess III

原文：http://www.cnblogs.com/sola1994/p/4151997.html

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