poj 2229

Sumsets Time Limit: 2000MS Memory Limit: 200000KTotal Submissions: 17820 Accepted: 6984

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

这一题想到了思路很简单，但是却很有意思。

题目大意是将一个数表示成二进制的和，问总共有多少种表达的方式。

思路是对于n为奇数的情况，因为拆开之后一定有个1，所以和n-1的情况相同。n为偶数的情况时，又分为两种情况，有1和无1，有1时和n-2相同（因为分出两个1），无1和n/2相同（因为所有的数都可以被2整除）。

#include <iostream> #include <cstdio> using namespace std; const int maxn = 1e6 + 4; const int mode = 1e9; int dp[maxn]; int main() { int n; dp[1] = 1; dp[2] = 2; scanf("%d", &n); for(int i = 3; i <= n; ++i) { if(i % 2 == 1) { dp[i] = dp[i - 1]; } else { dp[i] = (dp[i - 2] + dp[i / 2]) % mode; } } printf("%d\n", dp[n]); return 0; }