题目链接：Bear and Raspberry
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?
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 15 10 7 3 20
output
input
6 2100 1 10 40 10 40
output
97
input
3 01 2 3
output
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.
解题思路：题目实质是 给一序列，找出后一个与前一个差距最大的值，并且结果要减去C。直接贪心加暴力，先扫一遍， 找出相邻元素中后一元素和前一元素差值最大的ans，如果ans
AC代码：
#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint main(){ #ifdef sxk freopen("in.txt","r",stdin); #endif int n, c, a, b; while(scanf("%d%d",&n, &c)!=EOF) { int ans = 0; a = 0; for(int i=1; i>b; if(a-b > ans) ans = a-b; a = b; } if(ans > c) ans -= c; else ans = 0; cout <br> <br> <p></p> <p><br> </p> <p><br> </p> </time.h></stdlib.h></math.h></string></map></set></queue></vector></algorithm></iostream></string.h></stdio.h>
