POJ 2229 Sumsets（数位DP）-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_26658823/article/details/52107441

本文介绍了一种求解特定整数N的不同2的幂次和集数量的算法。通过动态规划方法，针对输入整数N（1≤N≤1,000,000），计算所有可能的表示方式，并输出结果的最后9位数字。

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Sumsets

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

Sample Output

题目大意：给出一个数，求出和等于给出数的不同集合数，集合里的数只能是2的k次幂。

解题思路：定义dp[i]为数字i的最多集合数，当i为奇数时，dp[i] = dp[i - 1]；当i为偶数时，dp[i - 1]中的每个集合加1得到dp[i]中所有含1的集合，而dp[i / 2]乘2得到dp[i]中只有偶数的集合。

代码如下：

#include <algorithm>
#include <cctype>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define EPS 1e-6
#define INF INT_MAX / 10
#define LL long long
#define MOD 1000000000
#define PI acos(-1.0)
using namespace std;

const int maxn = 1000005;

LL dp[maxn];

int main()
{
    int n;
    scanf("%d",&n);
    dp[1] = 1;
    for(int i = 2;i <= n;i++){
        if(i % 2){
            dp[i] = dp[i - 1] % MOD;
        }
        else{
            dp[i] = dp[i - 1] % MOD + dp[i >> 1] % MOD;
        }
    }
    printf("%lld\n",dp[n]);
    return 0;
}