Codeforces Round #226 (Div. 2) A. Bear and Raspberry

本文探讨了一只熊如何通过巧妙地借入并卖出蜂蜜桶来最大化获取覆盆子的数量。通过对不同日期蜂蜜价格的分析,找出最佳买卖时机,实现利益最大化。

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A. Bear and Raspberry
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following n days. According to the bear's data, on the i-th (1 ≤ i ≤ n) day, the price for one barrel of honey is going to is xikilos of raspberry.

Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for c kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day d (1 ≤ d < n), lent a barrel of honey and immediately (on day d) sell it according to a daily exchange rate. The next day (d + 1)the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day d + 1) give his friend the borrowed barrel of honey as well as c kilograms of raspberry for renting the barrel.

The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan.

Input

The first line contains two space-separated integers, n and c (2 ≤ n ≤ 100, 0 ≤ c ≤ 100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.

The second line contains n space-separated integers x1, x2, ..., xn (0 ≤ xi ≤ 100), the price of a honey barrel on day i.

Output

Print a single integer — the answer to the problem.

Sample test(s)
input
5 1
5 10 7 3 20
output
3
input
6 2
100 1 10 40 10 40
output
97
input
3 0
1 2 3
output
0
Note

In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.

In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97

绝逼一道水题。。。

简单地求最大值就可以了

因为没注意到题目开头中的仅讨论d天借入和d+1天还回的情况,把题目想复杂了,浪费了点时间

用了23分钟才搞定。。。大哭大哭

贴上代码:

#include <iostream>
#include <string>
#include <vector>
#define rep(i,j,k) for(int i=(j); i<k; i++)
#define long long ll
#define maxn 110
int a[maxn];
int b[10100];
using namespace std;
int main(void){
    int n,c;
    while(cin >> n >> c){
     for(int i=1; i<=n; ++i)
      cin >> a[i];
     int k = 1;
     for(int i=1; i<n; ++i)
       b[k++] = a[i]-a[i+1]-c;
     int imax = b[1];
     for(int i=2; i<k; ++i)
      if(b[i] > imax)
       imax = b[i];
     if(imax <= 0)
      cout << "0" << endl;
     else
      cout << imax << endl;
    } 
    return 0;
}


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