POJ 2229-Sumsets ( 基础DP)

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

看了题解才会的题0.0...神奇的DP..

本题分为两种情况, 奇数和偶数.

使用数组dp[] 记录有几种组合..

1 奇数 dp[n]=dp[n-1]..这种情况自己想的到..

2 偶数n :a. dp[n] 的一部分是 dp[n-2] , 即再加上两个1 的情况.. //分解式中有1的情况

                b. 另一部分是 dp[n/2], 即没有 1 的情况, 将 n/2 中的组合乘以 2 即可...//分解式中没有1的情况

15.5.28

重新DP，写出来了=v=，

CODE:

#include <iostream>#include<stdio.h>using namespace std;const int M=1000000000;int dp[1000005];int main(){    int n;    dp[0]=dp[1]=1;    scanf("%d",&n);    for(int i=2;i<=n;i++)    {        if(i%2==1) dp[i]=dp[i-1];        else dp[i]=(dp[i-2]+dp[i/2])%M;    }    printf("%d\n",dp[n]);    return 0;}