Atcoder 2152 Boxes and Candies（模拟 + 水题）

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该博客介绍了Atcoder上的题目C - Boxes and Candies，要求在满足相邻两个盒子糖果数量差不超过x的情况下，找出最少需要吃掉多少糖果。通过遍历并调整每个盒子的糖果数，可以找到解题策略。提供了样例输入和输出，以及简单的解题代码。

题目链接：http://abc048.contest.atcoder.jp/tasks/arc064_a?lang=en

C - Boxes and Candies

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

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.

Sample Input 1

Copy

3 3
2 2 2

Sample Output 1

Copy

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

Sample Input 2

Copy

6 1
1 6 1 2 0 4

Sample Output 2

Copy

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).

Sample Input 3

Copy

5 9
3 1 4 1 5

Sample Output 3

Copy

The objective is already achieved without performing operations.

Sample Input 4

Copy

2 0
5 5

Sample Output 4

Copy

All the candies need to be eaten.

Submit

题意：n个盒子，每个盒子里面都有a[i]个糖果，要求每个两个盒子糖果差不能超过x，问至少要吃去几个糖果

解析：遍历一遍就可以了，一直吃右侧盒子里的糖果

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<cmath>
#define N 100009
using namespace std;
const int INF = 0x3f3f3f3f;
int a[N];

int main()
{
    int n, x;
    scanf("%d%d", &n, &x);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    long long ans, now; ans = now = 0;
    for(int i = 1; i <= n; i++)
    {
        if(now + a[i] <= x) now = a[i];
        else ans += now + a[i] - x, now = a[i] - (now + a[i] - x);
    }
    printf("%lld\n", ans);
    return 0;
}