HDU4727：The Number Off of FFF

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
        

題意：隊列裡所有人進行報數，要找出報錯的那個人

思路：其實是很水的題，隻要找出序列中與錢一個人的數字差不是1的人即可，但是要注意的是這些人是從一個隊列中間截取下來的，是以很有可能第一個人就報錯了，一開始沒有考慮這種狀況是以出錯了

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int num[100005];

int main()
{
    int T,cas = 1,n,i,x,flag;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        x = -1;
        flag = -1;
        for(i = 0; i<n; i++)
            scanf("%d",&num[i]);
        for(i = 1;i<n;i++)
        {
            if(num[i]-num[i-1]!=1)
            flag = i+1;
        }
        if(flag==-1)
        flag = 1;
        printf("Case #%d: ",cas++);
        printf("%d\n",flag);
    }

    return 0;
}