poj 2229 Sumsets

http://poj.org/problem?id=2229

题意：把一个整数拆分为2的幂。

当n为奇数时dp
=dp[n-1];因为每总都加1，所以总数一样。当n为偶数时，分为有1和无1，有1.拆一个1，就变成奇数了，无1,就等于dp[n/2];

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 14180		Accepted: 5637

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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int m=1000000000;
int dp[1000000];
int main()
{
int n;
while(~scanf("%d",&n))
{
memset(dp,0,sizeof(dp));
int i;
dp[1]=1;
dp[2]=2;
for(i=3;i<=n;i++)
{
if(i%2==0)
{
dp[i]=dp[i/2]+dp[i-2];
dp[i]%=m;
}
else
{
dp[i]=dp[i-1];
dp[i]%=m;
}
}
printf("%d\n",dp
%m);

}
return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int m=1000000000;
int dp[1000000];
int main()
{
int n;
while(~scanf("%d",&n))
{
memset(dp,0,sizeof(dp));
int i;
dp[1]=1;
dp[2]=2;
for(i=3;i<=n;i++)
{
if(i%2==0)
{
dp[i]=dp[i/2]+dp[i-2];
dp[i]%=m;
}
else
{
dp[i]=dp[i-1];
dp[i]%=m;
}
}
printf("%d\n",dp
%m);

}
return 0;
}