POJ_2229_Sumsets_(动态规划)
2016-04-23 18:11
239 查看
描述
http://poj.org/problem?id=2229将一个数n分解为2的幂之和共有几种分法?
Sumsets
Time Limit: 2000MS | Memory Limit: 200000K | |
Total Submissions: 16207 | Accepted: 6405 |
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分析
对i讨论:
1.i是奇数:
分成的序列中必有1,所以可将i分为1+(i-1),所以f[i]=f[i-1];
2.i是偶数:
(1).分成的序列中有1:
同奇数,f[i]=f[i-1];
(2).分成的序列中没有1:
序列中的所有数都是2的倍数,那么任一种序列中的各个数/2,就得到了i/2的序列,那这种情况下,i的序列数就和i/2的序列数相同即f[i]=f[i/2];
综上:f[i]=f[i-1]+f[i/2];
#include<cstdio> const int maxn=1000005,mod=1e9; int n,f[maxn]; int main() { #ifndef ONLINE_JUDGE freopen("sum.in","r",stdin); freopen("sum.out","w",stdout); #endif scanf("%d",&n); f[1]=1; for(int i=2;i<=n;i++) { if(i&1) f[i]=f[i-1]; else f[i]=(f[i-1]+f[i/2])%mod; } printf("%d\n",f ); #ifndef ONLINE_JUDGE fclose(stdin); fclose(stdout); #endif return 0; }View Code
相关文章推荐
- 辗转相除法的证明
- Intellij idea断点 Debugger slow: Method breakpoints my dramatically slow down debugging
- Costume Party
- opencv学习记录1--基本概念(1)
- 走迷宫
- 树莓派 Linux备忘
- sublime 插件
- 热点推荐:秒杀系统架构分析与实战
- 如何让 帝国cms文章内容tags关键词设置调用方法以及tags静态化?
- ACM2-1013
- CodeForces - 420A (字符对称问题)
- linux shell脚本备份mysql数据库
- IDEA中maven如何将jar包导入本地的maven库
- LeetCode *** 222. Count Complete Tree Nodes
- 从eclipse到android studio
- Eclipse中的java、xml代码提示功能设置
- python中的json模块
- c# 获得文字的像素宽度
- linux下安装Apache(https) 服务器证书安装配置指南
- 极客WEB大前端专家级开发工程师培训视频教程