POJ 3427 Ecology tax

最新推荐文章于 2017-08-02 18:58:43 发布

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最新推荐文章于 2017-08-02 18:58:43 发布

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分类专栏： ACM_POJ_模拟题文章标签： tree output resources integer input logging

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本文介绍了一种通过调整树木生长周期来最小化木材浪费税收的方法。政府规定了木材运输的标准长度，公司需按浪费比例缴纳税款。文章提供了一个算法，用于计算达到最低税收所需的最短等待年数。

http://poj.org/problem?id=3427

Ecology tax

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3453		Accepted: 1777

Description

In a big and rich on natural resources country, the government started a campaign to control deforestation. In fact the government is not too interested in how many trees get fallen, but rather how effectively the wood is utilized. So a law was passed which requires every logging company to pay amount of money in proportion to amount of wood that it wastes during operation.

A felling quota on some territory was allotted to a company in this country. Company lorries may only transport logs of exactly L meters long. So when a tree gets sawed into logs, the remainder is wasted.

Trees in this country grow exactly 1 meter per year, so the company may decrease the amount of tax to be paid by simply waiting for some years. Your task is to determine the number of years needed to achieve smallest possible tax. If there is more than one answer, find minimal (earliest) one.

Input

Input file contains number of trees N, length of log L, followed by integers i₁ i₂ ... i_N — heights of each tree.

Constraints

1 ≤ N ≤ 30000, 1 ≤ L, i_k ≤ 30000

Output

Output file must contain single integer — number of years to wait.

Sample Input

Sample Input 1
3 1 
10 15 11
Sample Input 2
3 2
5 3 6

Sample Output

Sample Output 1
0
Sample Output 2
1

Hint

Bold texts appearing in the sample sections are informative and do not form part of the actual data.

Source

Northeastern Europe 2006, Far-Eastern Subregion

#include<iostream>
using namespace std;

int max(int a, int b){
return a > b ? a : b;
}

int main(){
   int tree, len;
    while(cin >> tree >> len){
        int hei, ans = 0;
        while(tree --){
           cin >> hei;
            if(hei % len == 0) continue;
            ans = max(ans, len - hei % len);
       }
       cout << ans << endl;
    }
    return 0;
}