POJ2229 Sumsets

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 19024		Accepted: 7431

Description

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

Source

USACO 2005 January Silver

——————————————————————————————————

题目的意思是给出一个数问把他变成若干个2^x的数累加，问有多少种不同情况

思路：方法一：dp，完全背包 +打表

方法二：分析可知对于第i项，i为奇数项就等于i-1项的值，i为偶数项就等于i-1项    加上i/2项的值（把i/2项每个数*2）

方法一：可能会超时，看判题机状态

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
 
using namespace std;
 
#define LL long long
const int INF = 0x3f3f3f3f;
const int mod=1000000000;
int dp[1000005];
 
int main()
{
    int n;
    memset(dp,0,sizeof dp);
    dp[0]=1;
    for(int i=0;i<=20;i++)
    {
        int k=pow(2,i);
        for(int j=k;j<1000005;j++)
        {
            dp[j]=(dp[j]+dp[j-k])%mod;
        }
    }
    while(~scanf("%d",&n)){
        printf("%d\n",dp[n]);
 
    }
 
    return 0;
}

　　

方法二：效率较高

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
 
using namespace std;
 
#define LL long long
const int INF = 0x3f3f3f3f;
const int mod=1000000000;
int dp[1000005];
 
int main()
{
    int n;
    memset(dp,0,sizeof dp);
    dp[1]=1;
        for(int j=2;j<1000005;j++)
        {
           if(j%2)
            dp[j]=dp[j-1]%mod;
           else
            dp[j]=(dp[j-1]+dp[j/2])%mod;
        }
    while(~scanf("%d",&n)){
        printf("%d\n",dp[n]);
 
    }
 
    return 0;
}