2019.9.27 split of natural numbers

Title Description

Given a natural number N, N required to be split into several positive integers the sum of formats in the adding operation may be repeated a number. Program Number 2147483648 seeking result of splitting mod. 1≤N≤4000. 

Entry

An integer n.

Export

The output of a number, the number of programs that is all 
because this number may be quite large, so you just output this number 2147483648 to the remainder mod. 

Sample input

7

Sample Output

14

prompt

Split 7 of a total of 14 cases 

7=1+6 

7=1+1+5 

7=1+1+1+4 

7=1+1+1+1+3 

7=1+1+1+1+1+2 

7=1+1+1+1+1+1+1 

7=1+1+1+2+2 

7=1+1+2+3 

7=1+2+4 

7=1+2+2+2 

7=1+3+3 

7=2+5 

7=2+2+3 

7=3+4 

---

Simple dp, just pay attention to enumerate items first and then to enumerate capacity.

 

#include<iostream>
#include<cstdio>
#include<cstring>
#define int long long
#define mod 2147483648
using namespace std;
int n,dp[100050];
signed main()
{
	scanf("%lld",&n);
	dp[0]=1;
	for(int i=1;i<n;i++)
	{
		for(int j=i;j<=n;j++)
		{
			dp[j]=(dp[j]+dp[j-i])%mod;
		}
	}
	printf("%lld",dp[n]);
	return 0;
}