一、題目

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問(wèn)題，先討論下起始情況：當(dāng)n = 1的時(shí)候，只有1種；當(dāng)n = 2的時(shí)候，有兩種；
• 當(dāng)n為奇數(shù)時(shí)，將n - 1的所有情形都加上1即可以得到n時(shí)的個(gè)數(shù)。dp[n] = dp[n - 1]；
• 當(dāng)n為偶數(shù)時(shí)，又分為兩種情況：若式子含有1，則這種情況的個(gè)數(shù)為dp[n - 1]；若式子不包含1，即全都是偶數(shù)，則將所有加數(shù)都除以2，可得到個(gè)數(shù)為dp[n / 2]的值；
• 題目中要求的輸出為實(shí)際值的最后九位數(shù)字，所以每次計(jì)算時(shí)要對(duì)1000000000取模。

三、代碼

#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; }

轉(zhuǎn)載于:https://www.cnblogs.com/CSLaker/p/7400179.html

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