poj 2229

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 15105		Accepted: 6017

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

 

 

 

有两种做法

第一种是比较暴力的完全背包:

　　　　设状态dp[i]为在总和为i的情况下符合题意的方法总数，因为n不超过2 ^ 20，所以可以把这20个数看成是物品，则可以用完全背包的方法写出状态方程

　　　　　　dp[i] = dp[i - (1 << j)] ; ((1 << j) <= n)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

const int maxn = 1e6 + 7;
const int mod = 1e9;
int dp[maxn];
int main() {
    int n;
    scanf("%d", &n);
    dp[0] = 1;
    for (int i = 0; 1 << i <= n; ++i) {
        for (int j = 1 << i; j <= n; ++j) {
            dp[j] = (dp[j] + dp[j - (1 << i)]) % mod;
        }
    }

    printf("%d
", dp[n]);


    return 0;
}

第二种是略有技巧性的递推：

　　　　 同样设状态dp[i]为在总和为i的情况下符合题意的方法总数，考虑n为奇数和n为偶数的两种情况。

　　　　　　当n为奇数时，容易知序列中必含有1

　　　　　　　　dp[i] = dp[i - 1];

　　　　　　当n为偶数时，若序列中有1则至少含有两个1，dp[i] += dp[i - 2];

　　　　　　　　　　　　  若序列中没有1，则容易知序列所有数都是偶数，则dp[i] += dp[i / 2];

　　　　　　　　　　　　两式相加得dp[i] = dp[i - 2] + dp[i / 2];

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

const int maxn = 1e6 + 7;
const int mod = 1e9;
int dp[maxn];
int main() {
    int n;
    scanf("%d", &n);
    dp[0] = 1;
    for (int i = 1; i <= n; ++i) {
        if (i % 2 == 1) {
            dp[i] = dp[i - 1];
        } else {
            dp[i] = (dp[i - 2] + dp[i / 2]) % mod;
        }
    }

    printf("%d
", dp[n]);


    return 0;
}