POJ2229 解题报告

Sumsets

Time Limit: 2000MSMemory Limit: 200000KTotal Submissions: 19360Accepted: 7559

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

下面写题解

dp[i]:数字i的最多组合数
若i是奇数：i的任意一个组合都包含1，所以dp[i] = dp[i-1]
若i是偶数：i-1是奇数，任意一个组合加1构成i的组合，除了这些组合之外，i的其他组合都不包含1，意味着全部是偶 数，则i/2的每个组合乘2可以构成i的一个组合，所以dp[i]=dp[i-1]+dp[1/2]

#include<iostream>#include<algorithm>using namespace std;const int maxn=1000000+100;const int MOD=1000000000;int ans[maxn];int main(){    int n;    ans[1]=1;    for(int i=2;i<maxn;i++)        if(i&1) ans[i]=ans[i-1];        else ans[i]=(ans[i-1]+ans[i>>1])%MOD;    while(cin>>n) cout<<ans[n]<<endl;    return 0;}