【DP】poj2229_Sumsets_201408041139

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Sumsets

Time Limit: 2000MS Memory Limit: 200000KTotal Submissions: 12692 Accepted: 5091

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

Source

USACO 2005 January Silver

题意：把一个整数划分成2的次幂相加的方案数（类似于整数划分）

解题思路：

可以把n分成奇数和偶数来分别对待：因为2^0=1,2^1=2,2^2=4......

1、n为奇数：ｎ的划分中肯定都要含１，所以　num[n] = num [n-1];　即把ｎ－１的划分方案每个加１

2、ｎ为偶数：可以分成两种情况来考虑：一、划分方案中含有１（那么１的个数肯定为偶数），（这时ｎ－１为奇数，ｎ－１的划分中都含１），所以划分方案＝num[n-1];

二、划分方案中不含１，那么划分中肯定都是２的倍数，（因为只有２︿０＝１），所以划分方案＝num[n／２];

　　即ｎ为偶数时　num[n] = num[n-1] + num[n/2] ;

附ＡＣ代码：

#include <stdio.h>#include <string.h>#include <stdlib.h>#define mod 1000000000int f(int n){    int i,j,k,*s;    s = (int *)malloc((n+1)*sizeof(int));    memset(s,0,4*(n+1));        s[1]=1;    for(i=2;i<=n;i++)    {        if(i&1)        s[i]=s[i-1]%mod;        else        {            s[i]=(s[i-1]+s[i/2])%mod;        }    }    return s[n]%mod;}int main(){    int n;    scanf("%d",&n);          printf("%d\n",f(n));    //system("pause");    return 0;}

体会：

没好好看题，第一眼看到划分序列，立马想到整数划分，然后按照那个思路写，然后也没写对

这道题运用ｄｐ的思想，然后根据题意，抓住２的次幂，很巧妙地解法

标签：ＤＰ

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