Codeforces 451B

最新推荐文章于 2024-04-25 23:38:56 发布

吹石柳生

最新推荐文章于 2024-04-25 23:38:56 发布

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分类专栏： ACM题目文章标签：模拟水题 codeforces

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本文探讨了一个编程问题，即在给定的数组中，通过仅反转一个子段来实现整数组的升序排列。文章详细介绍了解决思路、输入输出格式，并提供了实现代码，帮助读者理解如何通过数组操作实现特定的排序目标。

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Being a programmer, you like arrays a lot. For your birthday, your friends have given you an array a consisting of n distinct integers.

Unfortunately, the size of a is too small. You want a bigger array! Your friends agree to give you a bigger array, but only if you are able to answer the following question correctly: is it possible to sort the array a (in increasing order) by reversing exactly one segment of a? See definitions of segment and reversing in the notes.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 10⁵) — the size of array a.

The second line contains n distinct space-separated integers: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 10⁹).

Output

Print "yes" or "no" (without quotes), depending on the answer.

If your answer is "yes", then also print two space-separated integers denoting start and end (start must not be greater than end) indices of the segment to be reversed. If there are multiple ways of selecting these indices, print any of them.

Sample test(s)

Input

3
3 2 1

Output

yes
1 3

Input

4
2 1 3 4

Output

yes
1 2

Input

4
3 1 2 4

Output

no

Input

2
1 2

Output

yes
1 1

这道题用模拟就可以了，直接上代码了。

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 100010;
struct Node{
    int val,id;
    bool operator < (const Node& rhs) const{
        return val < rhs.val;
    }
}node[maxn];

int main(){
    int n,l = -1,r;
    bool flag = 1;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++){
        scanf("%d",&node[i].val);
        node[i].id = i;
    }
    sort(node+1,node+n+1);
    node[n+1].id = -1;
    for(int i = 1;i <= n;i++){
        if(node[i].id != i){
            l = i;
            break;
        }
    }
    if(l == -1){
        printf("yes\n1 1\n");
        return 0;
    }
    for(int i = l;i <= n;i++){
        if(node[i].id - node[i+1].id != 1){
            r = i;
            break;
        }
    }
    for(int i = r+1;i <= n;i++){
        if(node[i].id != i){
            flag = false;
            break;
        }
    }
    if(flag)    printf("yes\n%d %d\n",l,r);
    else        printf("no\n");
    return 0;
}