CF1442A Extreme Subtraction

Extreme Subtraction

题目描述

You are given an array $a$ of $n$ positive integers.

You can use the following operation as many times as you like: select any integer $1\le k\le n$ and do one of two things:

• decrement by one $k$ of the first elements of the array.
• decrement by one $k$ of the last elements of the array.

For example, if $n=5$ and $a=[3,2,2,1,4]$ , then you can apply one of the following operations to it (not all possible options are listed below):

• decrement from the first two elements of the array. After this operation $a=[2,1,2,1,4]$ ;
• decrement from the last three elements of the array. After this operation $a=[3,2,1,0,3]$ ;
• decrement from the first five elements of the array. After this operation $a=[2,1,1,0,3]$ ;

Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.

输入格式

The first line contains one positive integer $t$ ( $1\le t\le 30000$ ) — the number of test cases. Then $ t $ test cases follow.

Each test case begins with a line containing one integer $n$( $1\le n\le 30000$ ) — the number of elements in the array.

The second line of each test case contains $n$ integers $a_1\dots a_n$ ( $1\le a_i\le 10^6$ ).

The sum of $ n $ over all test cases does not exceed $30000$ .

输出格式

For each test case, output on a separate line:

• YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
• NO, otherwise.

The letters in the words YES and NO can be outputed in any case.

题面翻译

你有一个序列 $a$，你可以进行 $2$ 种操作：

• 选择前 $k$ 个数，将它们全部减 $1$
• 选择后 $k$ 个数，将它们全部减 $1$

$k$ 由你自己定，并且每次操作可以不同。
问是否可以把通过若干次操作整个序列变为全是 $0$ 的序列

本题多测，有 $t$ 组数据
$t\le 30000$，$\sum n\le 30000$，$a_i\le {10}^6$

样例 #1

样例输入 #1

4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10

样例输出 #1

YES
YES
NO
YES

---

思路：

将此题目转化成一个差分问题。要使数组中的全部数都为 $0$，那么差分数组也必须为 $0$ 且 $d_1=0$。

那么我们来看两种操作对于差分数组有何影响：

操作 $1$：d[1]−1且 d[i+1]+1。那么我们就可以凭借操作 $1$，将差分数组中的负数变成 $0$，同时减小 $d_1$。

操作 $2$：d[i+1]−1且 d[n+1]+1。那么我们就可以凭借操作 $2$，将差分数组中的正数变成 $0$，而 $d_{n+1}$ 为多少和我们并没有关系。

在最开始时 $d_1=a_1$，所以只要差分数组的负数和的绝对值小于等于 $a_1$,即输出 YES，否则输出 NO。

---

代码：

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+10;
const int INF=0x3f3f3f3f;
int T,n,a[N],b[N],s;
signed main(){
	cin>>T;
	while(T--){
		s=0;
		cin>>n;
		for(int i=1;i<=n;i++){
			a[i]=b[i]=0;
		}
		for(int i=1;i<=n;i++){
			cin>>a[i];
			b[i]=a[i]-a[i-1];
		}
		for(int i=2;i<=n;i++){
			if(b[i]<0){
				s-=b[i];
			}
		}
		if(s<=a[1]){
			cout<<"YES"<<endl;
		}else{
			cout<<"NO"<<endl;
		}
	}
	return 0;
}

完结撒花~