ACM思维题训练集合
The Little Elephant has got a problem — somebody has been touching his sorted by non-decreasing array a of length n and possibly swapped some elements of the array.
The Little Elephant doesn't want to call the police until he understands if he could have accidentally changed the array himself. He thinks that he could have accidentally changed array a, only if array a can be sorted in no more than one operation of swapping elements (not necessarily adjacent). That is, the Little Elephant could have accidentally swapped some two elements.
Help the Little Elephant, determine if he could have accidentally changed the array a, sorted by non-decreasing, himself.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the size of array a. The next line contains n positive integers, separated by single spaces and not exceeding 109, — array a.
Note that the elements of the array are not necessarily distinct numbers.
Output
In a single line print "YES" (without the quotes) if the Little Elephant could have accidentally changed the array himself, and "NO" (without the quotes) otherwise.
Examples
Input
2
1 2
Output
YES
Input
3
3 2 1
Output
YES
Input
4
4 3 2 1
Output
NO
Note
In the first sample the array has already been sorted, so to sort it, we need 0 swap operations, that is not more than 1. Thus, the answer is "YES".
In the second sample we can sort the array if we swap elements 1 and 3, so we need 1 swap operation to sort the array. Thus, the answer is "YES".
In the third sample we can't sort the array in more than one swap operation, so the answer is "NO".
题目中说原来的数列非递减，也就是非严格递增，显然原数列易得，排遍序即可，如果发生了一次交换至多有两个不一样。
#include <bits/stdc++.h>
using namespace std;
const int maxn=100055;
int a[maxn];
int b[maxn];
int main()
{
int n;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>a[i];
b[i]=a[i];
}
sort(b,b+n);
int cnt=0;
for(int i=0;i<n;i++)
{
if(a[i]!=b[i]) cnt++;
if(cnt==3) break;
}
if(cnt>2) cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}