Boxes and Candies(贪心)

Boxes and Candies

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

There are N boxes arranged in a row. Initially, the i-th box from the left contains ai candies.

Snuke can perform the following operation any number of times:

• Choose a box containing at least one candy, and eat one of the candies in the chosen box.

His objective is as follows:

• Any two neighboring boxes contain at most x candies in total.

Find the minimum number of operations required to achieve the objective.

Constraints

• 2≤N≤105
• 0≤ai≤109
• 0≤x≤109

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Input

The input is given from Standard Input in the following format:

N x
a1 a2 … aN

Output

Print the minimum number of operations required to achieve the objective.

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Sample Input 1

Copy

3 3
2 2 2

Sample Output 1

Copy

1

Eat one candy in the second box. Then, the number of candies in each box becomes (2,1,2).

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Sample Input 2

Copy

6 1
1 6 1 2 0 4

Sample Output 2

Copy

11

For example, eat six candies in the second box, two in the fourth box, and three in the sixth box. Then, the number of candies in each box becomes (1,0,1,0,0,1).

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Sample Input 3

Copy

5 9
3 1 4 1 5

Sample Output 3

Copy

0

The objective is already achieved without performing operations.

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Sample Input 4

Copy

2 0
5 5

Sample Output 4

Copy

10

All the candies need to be eaten.

//一道简单的贪心，我想得太复杂了。。。而且结果要用 long long 保存

 #include <stdio.h>

 int box[];
 int main()
 {
     int n,x;
     while (scanf("%d%d",&n,&x)!=EOF)
     {
         for (int i=;i<=n;i++)
             scanf("%d",&box[i]);
         long long all=;
         if (box[]>x)
         {
             all+=box[]-x;
             box[]=x;
         }
         for (int i=;i<=n;i++)
         {
             int k = box[i]+box[i-]-x;
             if (k>)
             {
                 all+=k;
                 box[i]-=k;
             }
         }
         printf("%lld\n",all);
     }
 }