A. Taming the Herd

本文解析了 CodeForces 上的一道经典问题——牧场逃脱,详细阐述了问题背景、解决思路及代码实现。文章指出,正确理解题意是关键,避免因语言障碍而误解题目的要求。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

http://codeforces.com/group/NVaJtLaLjS/contest/238204/problem/A

A. Taming the Herd

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Early in the morning, Farmer John woke up to the sound of splintering wood. It was the cows, and they were breaking out of the barn again!

Farmer John was sick and tired of the cows' morning breakouts, and he decided enough was enough: it was time to get tough. He nailed to the barn wall a counter tracking the number of days since the last breakout. So if a breakout occurred in the morning, the counter would be 00that day; if the most recent breakout was 33 days ago, the counter would read 33. Farmer John meticulously logged the counter every day.

The end of the year has come, and Farmer John is ready to do some accounting. The cows will pay, he says! But lo and behold, some entries of his log are missing!

Farmer John is confident that the he started his log on the day of a breakout. Please help him determine, out of all sequences of events consistent with the log entries that remain, the minimum and maximum number of breakouts that may have take place over the course of the logged time.

Input

The first line contains a single integer NN (1≤N≤1001≤N≤100), denoting the number of days since Farmer John started logging the cow breakout counter.

The second line contains NN space-separated integers. The iith integer is either −1−1, indicating that the log entry for day ii is missing, or a non-negative integer aiai (at most 100100), indicating that on day ii the counter was at aiai.

Output

If there is no sequence of events consistent with Farmer John's partial log and his knowledge that the cows definitely broke out of the barn on the morning of day 11, output a single integer −1−1. Otherwise, output two space-separated integers mm followed by MM, where mm is the minimum number of breakouts of any consistent sequence of events, and MM is the maximum.

Example

input

Copy

4
-1 -1 -1 1

output

Copy

2 3

Note

In this example, we can deduce that a breakout had to occur on day 3. Knowing that a breakout also occurred on day 1, the only remaining bit of uncertainty is whether a breakout occurred on day 2. Hence, there were between 2 and 3 breakouts in total.

题目分析:

读错题导致一直WA。没有理解什么情况输出-1,若给出数据中出现矛盾输出-1.我以为是全为-1时,输出-1.英语水平要待提高。

题读对后,题目不难。

代码:

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int N=1000;
int a[N],b[N],vis[N];int n;
int main()
{
	cin>>n;int flag=0;
	for(int i=1;i<=n;i++)cin>>a[i];
	memset(b,-1,sizeof(b));
	for(int i=0;i<=n;i++)vis[i]=1;
	for(int i=2;i<=n;i++)
	{
		if(a[i]>-1){
			if(b[i-a[i]]==0)flag=1;else b[i-a[i]]=1;
			for(int j=i-a[i]+1;j<=i;j++){
				if(b[j]==1)flag=1;
				else b[j]=0;
			}
		}		
	}
	b[1]=1;int c1=0,c2=0;
	for(int i=1;i<=n;i++){
		if(b[i]==1)c1++;
		if(b[i]==-1)c2++;
	}
	if(flag)puts("-1");
	else printf("%d %d\n",c1,c1+c2);
	
	
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值