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Codeforces Round #713 (Div. 3)-A. Spy Detected!

传送门 Time Limit: 2 seconds Memory Limit: 256 megabytes

Problem Description

You are given an array $a$ consisting of $n$ ( $n \ge 3$ ) positive integers. It is known that in this array, all the numbers except one are the same (for example, in the array $[4, 11, 4, 4]$ all numbers except one are equal to $4$ ).

Print the index of the element that does not equal others. The numbers in the array are numbered from one.

Input

The first line contains a single integer $t$ ( $1 \le t \le 100$ ). Then $t$ test cases follow.

The first line of each test case contains a single integer $n$ ( $3 \le n \le 100$ ) — the length of the array $a$ .

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $1 \le a_i \le 100$ ).

It is guaranteed that all the numbers except one in the $a$ array are the same.

Output

For each test case, output a single integer — the index of the element that is not equal to others.

Sample Input

4
4
11 13 11 11
5
1 4 4 4 4
10
3 3 3 3 10 3 3 3 3 3
3
20 20 10

Sample Onput

题目大意

给你 $n(n\geq3)$ 个数，其中只有一个数字与众不同。让你给出这个与众不同的数字下标。

解题思路

先看数据范围， $t<100,n<100,a_i<100$ ，所以暴力就够了

一个很笨的方法就是输入数据之后，把每个数出现的次数+1，然后遍历一遍，得到出现一次的那个数。再数组一遍数组，得到这个数的下标。

AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
int a[200];//记录题目中给的数组
int main()
{
    int N;
    cin>>N;
    while(N--)//N组测试样例
    {
        int n;
        cd(n);//scanf("%d", &n);
        int sum[200]={0};//记录出现的次数，sum[i]代表i出现的次数
        fi(i,0,200)//for(int i=0;i<200;i++)
            sum[i]=0;//初始值为0
        fi(i,0,n)
        {
            cd(a[i]);//输入a[i]
            sum[a[i]]++;//a[i]出现的次数++
        }
        int diff=0;//不同的那个数（只出现一次的那个数）
        fi(i,1,200)
        {
            if(sum[i]==1)//这个数只出现了一次
            {
                diff=i;
                break;
            }
        }
        // dbg(diff);
        fi(i,0,200)
        {
            if(a[i]==diff)//如果这个数就是只出现一次的那个数
            {
                printf("%d\n",i+1);//输出
                break;//结束循环
            }
        }
    }
    return 0;
}

同步发文于我的CSDN，原创不易，转载请附上原文链接哦~
Tisfy：letmefly.blog.csdn.net/article/det…

Codeforces Round #713 (Div. 3)-A. Spy Detected!-题解