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Codeforces Round #201 (Div. 2) B. Fixed Points

2013年02月06日 ⁄ 综合 ⁄ 共 1521字 ⁄ 字号 评论关闭
B. Fixed Points
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A permutation of length n is an integer sequence such that each integer from 0 to (n - 1) appears
exactly once in it. For example, sequence [0, 2, 1] is a permutation of length 3 while both [0, 2, 2] and [1, 2, 3] are
not.

A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer i is
a fixed point of permutation a0, a1, ..., an - 1 if
and only if ai = i.
For example, permutation [0, 2, 1] has 1 fixed point and permutation [0, 1, 2] has 3 fixed
points.

You are given permutation a. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in
the resulting permutation. Note that you are allowed to make at most one swap operation.

Input

The first line contains a single integer n (1 ≤ n ≤ 105).
The second line contains n integers a0, a1, ..., an - 1 —
the given permutation.

Output

Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation.

Sample test(s)
input
5
0 1 3 4 2
output
3
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define M 1000500
int prime[M];
int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF){

        int ans=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&prime[i]);
            if(prime[i]!=i)
            ans++;
        }
		if(ans==0)
		{
			printf("%d\n",n);
			continue;
		}
        bool flag=false;
        for(i=0;i<n;i++){
            if(prime[i]!=i){
               if(prime[prime[i]]==i){
                   //printf("%d ",i);
                    flag=true;
                    break;
                }
            }
        }
        if(flag){
            printf("%d\n",n-ans+2);
        }
        else
        printf("%d\n",n-ans+1);
    }

    return 0;
}

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