17 small A's candy, 17 candy

17 small A's candy, 17 candy
Description

John has N candy boxes and I has A [I] Candy.

Mr. A can eat one of the boxes of candy each time. He wants to know that to add any two adjacent boxes, there will be only x or less candy, you must eat at least a few pieces of sugar.

Input/Output Format

Input Format:

 

Enter N and x in the first line.

N integers in the second row, a [I].

 

Output Format:

 

The minimum number of sweets to be eaten.

 

Input and Output sample input sample #1:

3 32 2 2

Output sample #1:

1

Input example #2:

6 11 6 1 2 0 4

Output sample #2:

11

Input example #3:

5 93 1 4 1 5

Output sample #3:

0

Description

Example 1

Eat the candy in the second box.

Example 2

The second box eats 6, the fourth box eats 2, and the sixth box eats 3.

30% of test data, 2 <= N <= 20, 0 <= a [I], x <= 100

70% of test data, 2 <= N <= a [I], x <= 10 ^ 5

100% of test data, 2 <= N <= 10 ^ 5, 0 <= a [I], x <= 10 ^ 9

Mark

 

Mmp

Start

P = p-a [I + 1];
A [I + 1] = 0;

The write is reversed, so every time p is equal to no subtraction ,,

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 #define lli long long int  8 using namespace std; 9 void read(lli &n)10 {11     char c='+';lli x=0;bool flag=0;12     while(c<'0'||c>'9')13     {c=getchar();if(c=='-')flag=1;}14     while(c>='0'&&c<='9')15     {x=x*10+(c-48);c=getchar();}16     flag==1?n=-x:n=x;17 }18 lli a[100001];19 lli n,m;20 lli ans=0;21 int main()22 {23     read(n);read(m);24     for(lli i=1;i<=n;i++)25         read(a[i]);26     for(lli i=1;i<=n;i++)27     {28         if(a[i]+a[i+1]>m)29         {30             lli p=a[i]+a[i+1];31             if(a[i+1]>=(p-m))32             {33                 a[i+1]-=(p-m);34                 ans+=p-m;35             }36             else 37             {38                 ans+=a[i+1];39                 p=p-a[i+1];40                 a[i+1]=0;41                 a[i]-=(p-m);42                 ans+=(p-m);43             }44         }45     }46     printf("%lld",ans);47     return 0;48 }