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Sumsets【递推】

程序员文章站 2022-07-05 14:48:03
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Sumsets

 POJ - 2229 

Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7: 

1) 1+1+1+1+1+1+1 
2) 1+1+1+1+1+2 
3) 1+1+1+2+2 
4) 1+1+1+4 
5) 1+2+2+2 
6) 1+2+4 

Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000). 

Input

A single line with a single integer, N.

Output

The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).

Sample Input

7

Sample Output

6

题目大意:输入一个数n,输出这个数由2的幂次组成的情况数。

解决方法:通过列举1-10的结果,可以推出当n为偶数时dp[i]=dp[i-1]+dp[i/2]。

AC代码:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <utility>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e6 + 5;
const ll mod = 1e9;
ll dp[maxn];
int main() 
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ios::sync_with_stdio(0),cin.tie(0);
    dp[1]=1;
    rep(i,2,1000000) {
    	if(i%2==0)
    		dp[i]=(dp[i-1]+dp[i/2])%mod;
    	else
    		dp[i]=dp[i-1];
    }
    int n;
    cin>>n;
    cout<<dp[n]<<endl;
    return 0;
}

 

相关标签: 递推