Boxes and Candies

问题 G: Boxes and Candies

时间限制: 1 Sec  内存限制: 128 MB
[提交] [状态]

题目描述

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

输入

The input is given from Standard Input in the following format:
N x
a1 a2 … aN

输出

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

样例输入 Copy

3 3
2 2 2

样例输出 Copy

1

提示

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

问题大意：给定序列，可以一次改变一个值，使得任意相邻的两个之和小于等与k；

问题解析：先两个两个的改变，一定先改变两者后面的那个，因为后面的那个对后面的有贡献（比如说a1,a2,a3,先遍历a1和a2，判断是否符合要求，如果不符合要求，要先减少a2,因为a2还对a1有贡献）,注意不能让a2变成负数

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')f=-1,c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return f*x;}
typedef long long ll;
const int maxn=5e5+10;
const int M=1e7+10;
const int INF=0x3f3f3f3f;
ll n,x;
ll a[maxn];
void inint(){
    cin>>n>>x;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }    
}
int main(){
    inint();
    ll sum=0; 
    for(int i=1;i<n;i++){
        if(a[i]+a[i-1]>x){
            if(x-a[i-1]>=0){
                sum+=a[i]-(x-a[i-1]);
                a[i]=(x-a[i-1]);
            }
            else{
                sum+=(a[i]+a[i-1]-x);
                a[i]=0;
            }
        }
     }
    printf("%lld",sum);
    return 0;
}