CodeForces-1140A-Detective Book

本文介绍了一个简单的算法题目,通过遍历一次来计算阅读一本书所需的天数,书中每页包含一个谜团,谜团的解答在后续某一页中。文章提供了一个C++实现示例。

题目:

Description:

Ivan recently bought a detective book. The book is so interesting that each page of this book introduces some sort of a mystery, which will be explained later. The ii-th page contains some mystery that will be explained on page aiai (ai≥iai≥i).

Ivan wants to read the whole book. Each day, he reads the first page he didn't read earlier, and continues to read the following pages one by one, until all the mysteries he read about are explained and clear to him (Ivan stops if there does not exist any page ii such that Ivan already has read it, but hasn't read page aiai). After that, he closes the book and continues to read it on the following day from the next page.

How many days will it take to read the whole book?

Input

The first line contains single integer nn (1≤n≤1041≤n≤104) — the number of pages in the book.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (i≤ai≤ni≤ai≤n), where aiai is the number of page which contains the explanation of the mystery on page ii.

Output

Print one integer — the number of days it will take to read the whole book.

Example

Input

9
1 3 3 6 7 6 8 8 9

Output

4

Note

Explanation of the example test:

During the first day Ivan will read only the first page. During the second day Ivan will read pages number 22 and 33. During the third day — pages 44-88. During the fourth (and the last) day Ivan will read remaining page number 99.

题意分析:

水题吧,一遍for循环

#include<iostream>
#include<cstdio> 
#include<cstring>
#define N 10005
using namespace std;
int n,ans=0,now=0;
int a[N];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
	{
		if(a[i]>now)
		{
			now=a[i];
		}
		if(a[i]==now&&a[i]==i)ans++;
	}
	printf("%d",ans);
	return 0;
}

 

### 关于 Codeforces Problem 1802A 目前提供的引用内容并未涉及 Codeforces 编号为 1802A 的题目详情或解决方案[^1]。然而,基于常见的竞赛编程问题模式以及可能的解决方法,可以推测该类题目通常围绕算法设计、数据结构应用或者特定技巧展开。 如果假设此题属于典型的算法挑战之一,则可以从以下几个方面入手分析: #### 可能的方向一:字符串处理 许多入门级到中级难度的问题会考察字符串操作能力。例如判断子串是否存在、统计字符频率或是执行某种转换逻辑等。以下是 Python 中实现的一个简单例子用于演示如何高效地比较两个字符串是否相匹配: ```python def are_strings_equal(s1, s2): if len(s1) != len(s2): return False for i in range(len(s1)): if s1[i] != s2[i]: return False return True ``` #### 方向二:数组与列表的操作 另一常见主题是对整数序列进行各种形式上的变换或者是查询最值等问题。下面给出一段 C++ 程序片段来展示快速寻找最大元素位置的方法: ```cpp #include <bits/stdc++.h> using namespace std; int main(){ int n; cin >> n; vector<int> a(n); for(auto &x : a){ cin>>x; } auto max_it = max_element(a.begin(),a.end()); cout << distance(a.begin(),max_it)+1; // 输出索引加一作为答案 } ``` 由于具体描述缺失,在这里仅提供通用框架供参考。对于确切解答还需要访问实际页面获取更多信息后再做进一步探讨[^3]。
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