POJ 2229 Sumsets————（完全背包 or 规律）

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mb634e5c39919de 2022-10-18 16:34:16 博主文章分类：动态规划 ©著作权

文章标签 #include 完全背包 i++ 文章分类 OpenStack 云计算

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Sumsets
Time Limit: 2000MS Memory Limit: 200000K
Total Submissions: 23184 Accepted: 8888
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

将一个数划分成2的幂之和，一共有多少种划分方式
1 完全背包的思想
2 找规律

1.完全背包

/*
    利用完全背包的思想来解决 
*/
#include<cstdio> 
#include<cstring>
using namespace std;
const int MAXN=1e7+7;
const int mod=1e9;
int c[30];
int dp[MAXN];
int n,m;

int main() 
{
    int m=1;
    c[1]=1;
    while(1)
    {
        m++;
        c[m]=c[m-1]*2;
        if(c[m]>MAXN)   break;
    }
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=1;i<m;i++)
        {
            if(c[i]>n)  break;
            for(int j=c[i];j<=n;j++)
            {
                dp[j]+=dp[j-c[i]];;
                if(dp[j]>mod)   dp[j]%=mod;
            }
        }       
        printf("%d\n",dp[n]);
    }
    return 0;
}

#include<cstdio>
using namespace std;
const int mod=1e9;
const int MAXN=1e7+7;
int f[MAXN];

int main() 
{
    f[0]=1;
    f[1]=1;
    for(int i=2;i<MAXN;i++)
        if(i&1) f[i]=f[i-1];
        else    f[i]=f[i-1]%mod+f[i>>1]%mod;

    int n;
    while(~scanf("%d",&n)) 
        printf("%d\n",f[n]%mod);
    return 0;
}