【动态规划】POJ-2229

一、题目

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

二、思路&心得

• DP问题，先讨论下起始情况：当n = 1的时候，只有1种；当n = 2的时候，有两种；
• 当n为奇数时，将n - 1的所有情形都加上1即可以得到n时的个数。dp[n] = dp[n - 1]；
• 当n为偶数时，又分为两种情况：若式子含有1，则这种情况的个数为dp[n - 1]；若式子不包含1，即全都是偶数，则将所有加数都除以2，可得到个数为dp[n / 2]的值；
• 题目中要求的输出为实际值的最后九位数字，所以每次计算时要对1000000000取模。

三、代码

#include<stdio.h>
const int MAX_N = 1000005;
const int mod  = 1000000000;
int dp[MAX_N];
int f(int n) {
	for (int i = 3; i <= n; i ++) {
		if (i & 1) dp[i] = dp[i - 1];
		else dp[i] = (dp[i - 1] + dp[i / 2] ) % mod;
	}
	return dp[n];
}
int main() {
	int N;
	dp[1] = 1, dp[2] = 2;
	while (~scanf("%d", &N)) {
		printf("%d\n", f(N));
	}
	return 0;
}