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POJ 2229 Sumsets

时间:2015-03-08 13:00:11      阅读:119      评论:0      收藏:0      [点我收藏+]

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Sumsets

Time Limit: 2000ms
Memory Limit: 200000KB
This problem will be judged on PKU. Original ID: 2229
64-bit integer IO format: %lld      Java class name: Main
 
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

 
解题:背包。。。类比下就看出来了。。。
技术分享
 1 /*
 2 @author: Lev
 3 @date:
 4 */
 5 #include <iostream>
 6 #include <cstdio>
 7 #include <cmath>
 8 #include <cstring>
 9 #include <string>
10 #include <cstdlib>
11 #include <algorithm>
12 #include <map>
13 #include <set>
14 #include <queue>
15 #include <climits>
16 #include <deque>
17 #include <sstream>
18 #include <fstream>
19 #include <bitset>
20 #include <iomanip>
21 #define LL long long
22 #define INF 0x3f3f3f3f
23 
24 using namespace std;
25 const int maxn = 1000010;
26 int dp[maxn],c[25] = {1};
27 int main(){
28     int x;
29     for(int i = 1; i < 25; ++i)
30         c[i] = c[i-1]<<1;
31     while(~scanf("%d",&x)){
32         memset(dp,0,sizeof(dp));
33         dp[0] = 1;
34         for(int i = 0;c[i] <= x; ++i){
35             for(int j = c[i]; j <= x; ++j)
36             dp[j] = (dp[j] + dp[j-c[i]])%1000000000;
37         }
38         printf("%d\n",dp[x]);
39     }
40     return 0;
41 }
View Code

 

POJ 2229 Sumsets

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原文地址:http://www.cnblogs.com/crackpotisback/p/4321530.html

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