Poj 2229 (dp)

Source: Internet
Author: User
Sumsets
Time Limit:2000 MS   Memory Limit:200000 K
Total Submissions:10974   Accepted:4419

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 I started to think of as a full backpack, logn items, so the complexity of nlogn. The results are infinite. TLE is speechless. Then I searched for the question and found O (n) practices, which were quite clever. When n is an odd number, dp [n] = dp [n-1], because it must contain 1, therefore, each addition of 1 in dp [n-1] corresponds to dp [n] One by one. When n is an even number, it either contains at least 2 1, or does not include 1, if the value is not 1, the number of methods is to multiply each item of dp [n/2] by the value of 2. So dp [n] = dp [N-2] + dp [n/2]. Finally, the nlogn solution can be used. I just got stuck in two seconds to change the mod operation to subtraction. O (nlogn)
int dp[25][maxn];void solve(){    memset(dp,0,sizeof(dp));    for(int i = 0;i < 25;i++) dp[i][1] = dp[i][0] = 1;    for(int i = 1;i < 25;i++){        for(int j = 2;j < maxn;j++){            int v = 1 << (i-1);            if(j >= v) dp[i][j] = (LL(dp[i][j-v]) + dp[i-1][j]);            else dp[i][j] = dp[i-1][j];            while(dp[i][j] > Mod) dp[i][j] -= Mod;        }    }}int main(){    int n;    solve();    while(cin >> n){        cout << dp[24][n] << endl;    }    return 0;}

O (n)

LL dp[maxn];void solve(){    dp[0] = 1;    for(int i = 1;i < maxn;i++){        if(i%2 == 1) dp[i] = dp[i-1];        else dp[i] = (dp[i-2] + dp[i/2]) % Mod;    }}int main(){    int n;    solve();    while(scanf("%d",&n) != EOF){        printf("%d\n",dp[n]);    }    return 0;}

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.