# Codeforces Release B-Fixed Points

Source: Internet
Author: User
Question: Given a sequence, there is an operation: the positions of two numbers are interchangeable. Q: Can I perform an operation at most once? What is the maximum number of [element position I] in a sequence that is equal to [element ai?

Based on the question, the maximum number is:

[The number of equal [element position I] and [element ai] before the operation]+ 【Swap two [element position I] and [element ai] elements that are not equal to the number of [element position I] and [element ai ].

For example:

0 1 2 4 3

This sequence is used to replace the following two elements so that each[Element position I] is equal to [element ai].

That is to say, you don't have to consider it when switching.[Element position I] is equal to [element ai].

Direct consideration[Element position I] is not equal to [element ai]Can one or two elements be swapped once?[Element position I] is equal to [element ai].

If one element exists after replacement[Element position I] is equal to [element ai]

So the maximum number=[Before the operation, the number of equal [element position I] and [element ai] is + 1. If two elements exist, the number is + 2.

Next we can use map to solve this problem.

`#include <stdio.h>#include <map>using namespace std;int main(){    int n, ans;    while(~scanf("%d", &n))    {        map<int, int> mm;        ans = 0;        for(int i = 0; i < n; i++)        {            int temp;            scanf("%d", &temp);            if(temp == i)            {                ++ans;            }            else            {                mm[i] = temp;            }        }        map<int, int>::iterator it = mm.begin();        for(;it != mm.end();it++)        {            if(mm.find((*it).second) != mm.end() && mm[(*it).second] == (*it).first)            {                ++ans;                break;            }        }        if(ans != n)            ++ans;        printf("%d\n", ans);    }    return 0;}`

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