Codeforces Beta Round #25 (Div. 2)--A. IQ test

Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given n numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given n numbers finds one that is different in evenness.

Input

The first line contains integer n (3 ≤ n ≤ 100) — amount of numbers in the task. The second line contains n space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.

Output

Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.

Sample test(s)

input
5
2 4 7 8 10

output
3

input
4
1 2 1 1

output
2

解题报告：水题。。

。给你一列数，当中仅仅有一个数的奇偶性和其它数不同，输出它所在的位置。直接暴力扫一遍，记录最后一个奇数和偶数的位置，并记录奇数和偶数的个数。就可以。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

int a[105];

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int ans1 = 0, ans2 = 0, x, y;
        for(int i=0; i<n; i++){
            cin>>a[i];
            if(a[i]&1) {
                ans1 ++;
                x = i+1;
            }
            else{
                ans2 ++;
                y = i+1;
            }
        }
        if(ans1>ans2) cout<<y<<endl;         //推断条件也可换成ans1 == 1
        else cout<<x<<endl;
    }
    return 0;
}