HDU4727-The Number Off of FFF

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本文介绍了一道算法题目，需要找出连续队列中唯一报数错误的士兵位置。通过对比每个士兵报出的数字与前一名士兵数字的差值来判断是否报错。

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The Number Off of FFF

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3845 Accepted Submission(s): 1277

Problem Description

X soldiers from the famous " *FFF* army" is standing in a line, from left to right.
You, as the captain of *FFF*, decides to have a "number off", that is, each soldier, from left to right, calls out a number. The first soldier should call "One", each other soldier should call the number next to the number called out by the soldier on his left side. If every soldier has done it right, they will call out the numbers from 1 to X, one by one, from left to right.
Now we have a continuous part from the original line. There are N soldiers in the part. So in another word, we have the soldiers whose id are between A and A+N-1 (1 <= A <= A+N-1 <= X). However, we don't know the exactly value of A, but we are sure the soldiers stands continuously in the original line, from left to right.
We are sure among those N soldiers, exactly one soldier has made a mistake. Your task is to find that soldier.

Input

The rst line has a number T (T <= 10) , indicating the number of test cases.
For each test case there are two lines. First line has the number N, and the second line has N numbers, as described above. (3 <= N <= 10 ⁵)
It guaranteed that there is exactly one soldier who has made the mistake.

Output

For test case X, output in the form of "Case #X: L", L here means the position of soldier among the N soldiers counted from left to right based on 1.

Sample Input

  
   2
3
1 2 4
3
1001 1002 1004

Sample Output

  
   Case #1: 3
Case #2: 3

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

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zhuyuanchen520

题意：队列里所有人进行报数，要找出报错的那个人

解题思路：只要找出序列中与钱一个人的数字差不是1的人即可，但是这些人是从一个队列中间截取下来的，所以很有可能第一个人就报错了

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <functional>
#include <climits>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[100090],ans;

int main()
{
    int n,t;
    int cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++) scanf("%d",&a[i]);
        int ans=-1;
        for(int i=3; i<=n; i++)
        {
            if(a[i]-a[i-1]!=1)
            {
                ans=i;
                break;
            }
        }
        if(ans==-1) ans=1;
        printf("Case #%d: %d\n",cas++,ans);
    }
    return 0;
}