简单dp算法——Sumsets

B - Sumsets
点击打开链接 http://acm.hust.edu.cn/vjudge/contest/123760#problem/B

Crawling in process... Crawling failed Time Limit:2000MS     

Memory Limit:200000KB     64bit IO Format:%lld & %llu 

Submit Status

Description

Input

Output

Sample Input

Sample Output

Hint

Description

Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she produces as much milk as possible.

Farmer John has a list of M (1 ≤ M ≤ 1,000) possibly overlapping intervals in which he is available for milking. Each interval i has a starting hour (0 ≤ starting_houriN), an ending hour (starting_houri < ending_houriN), and a corresponding efficiency (1 ≤ efficiencyi ≤ 1,000,000) which indicates how many gallons of milk that he can get out of Bessie in that interval. Farmer John starts and stops milking at the beginning of the starting hour and ending hour, respectively. When being milked, Bessie must be milked through an entire interval.

Even Bessie has her limitations, though. After being milked during any interval, she must rest R (1 ≤ RN) hours before she can start milking again. Given Farmer Johns list of intervals, determine the maximum amount of milk that Bessie can produce in the N hours.

Input

* Line 1: Three space-separated integers: N, M, and R
* Lines 2..M+1: Line i+1 describes FJ's ith milking interval withthree space-separated integers: starting_houri , ending_houri , and efficiencyi

Output

* Line 1: The maximum number of gallons of milk that Bessie can product in the N hours

Sample Input

12 4 2
1 2 8
10 12 19
3 6 24
7 10 31

Sample Output

43

Hint

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

解题思路:
a[n]为和为 n 的种类数;
 根据题目可知,加数为2的N次方,即 n 为奇数时等于它前一个数 n-1 的种类数 a[n-1] ,若 n 为偶数时分加数中有无 1 讨论,即关键是对 n 为偶数时进行讨论:
 1.n为奇数,a[n]=a[n-1]
 2.n为偶数:
 (1)如果加数里含1,则一定至少有两个1,即对n-2的每一个加数式后面 +1+1,总类数为a[n-2];
 (2)如果加数里没有1,即对n/2的每一个加数式乘以2,总类数为a[n/2];
 所以总的种类数为:a[n]=a[n-2]+a[n/2];

代码实现:
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
int a[1000001];

int main()
{
    int   n;
    cin >> n;
    int i;
    a[1] = 1; a[2] = 2;
    for( i=3; i<=n; i++ )
    {
        if( i%2 == 1)
            a[i] = a[i-1];
        else
            a[i] = a[i-2] + a[i/2];
        a[i] = a[i] % 1000000000;
    }
    cout << a[n] <

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