CF - 451B. Sort the Array 排序+類比

1.題目描述：

B. Sort the Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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. Examples input

33 2 1

output

yes1 3

input

42 1 3 4

output

yes1 2

input

43 1 2 4

output

no

input

21 2

output

yes1 1

Note

Sample 1. You can reverse the entire array to get [1, 2, 3], which is sorted.

Sample 3. No segment can be reversed such that the array will be sorted.

Definitions

A segment [l, r] of array a is the sequence a[l], a[l + 1], ..., a[r].

If you have an array a of size n and you reverse its segment [l, r], the array will become:

a[1], a[2], ..., a[l - 2], a[l - 1], a[r], a[r - 1], ..., a[l + 1], a[l], a[r + 1], a[r + 2], ..., a[n - 1], a[n].

2.題意概述：

給你一段數，要你判斷是否能夠通過逆序一段字區間使得數嚴格不遞減

3.解題思路：

1、從左至右遍曆，第一次出現比前一個數小的下標；
2、然後從右往左遍曆，第一次出現比後一個數小的下標；
3、將這段區間逆序，然後判斷整個序列是否單調遞增

4.AC代碼：

#include <stdio.h>#include <algorithm>#define maxn 100100using namespace std;int a[maxn];int main(){int n;while (scanf("%d", &n) != EOF){int flag = 0, x = 0, y = 0;for (int i = 0; i < n; i++)scanf("%d", &a[i]);for (int i = 0; i < n - 1; i++)if (a[i] > a[i + 1]){x = i;break;}for (int i = n - 1; i > 0; i--)if (a[i] < a[i - 1]){y = i;break;}reverse(a + x, a + y + 1);for (int i = 0; i < n - 1; i++){if (a[i] > a[i + 1]){flag = 1;break;}}if (flag)puts("no");else{printf("yes\n%d %d\n", x + 1, y + 1);}}return 0;}