zoukankan      html  css  js  c++  java
  • HDU 4727 The Number Off of FFF (水题)

    The Number Off of FFF

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 78    Accepted Submission(s): 36


    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 <= 105)
    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
     
    Recommend
    zhuyuanchen520
     
     1 /* ***********************************************
     2 Author        :kuangbin
     3 Created Time  :2013-9-11 12:08:15
     4 File Name     :2013-9-111012.cpp
     5 ************************************************ */
     6 
     7 #include <stdio.h>
     8 #include <string.h>
     9 #include <iostream>
    10 #include <algorithm>
    11 #include <vector>
    12 #include <queue>
    13 #include <set>
    14 #include <map>
    15 #include <string>
    16 #include <math.h>
    17 #include <stdlib.h>
    18 #include <time.h>
    19 using namespace std;
    20 int a[100010];
    21 int main()
    22 {
    23     //freopen("in.txt","r",stdin);
    24     //freopen("out.txt","w",stdout);
    25     int T;
    26     int n;
    27     scanf("%d",&T);
    28     int iCase = 0;
    29     while(T--)
    30     {
    31         iCase++;
    32         scanf("%d",&n);
    33         for(int i = 0;i < n;i++)
    34             scanf("%d",&a[i]);
    35         int ans = 1;
    36         for(int i = 1;i < n;i++)
    37             if(a[i] != a[i-1] + 1)
    38             {
    39                 ans = i+1;
    40                 break;
    41             }
    42         printf("Case #%d: %d
    ",iCase,ans);
    43     }
    44     return 0;
    45 }
  • 相关阅读:
    【左偏树+贪心】BZOJ1367-[Baltic2004]sequence
    【裸裸的左偏树】BZOJ1455-罗马游戏
    【贪心+Treap】BZOJ1691-[Usaco2007 Dec]挑剔的美食家
    【贪心】BZOJ3668-[NOI2014]起床困难综合症
    【多重背包小小的优化(。・∀・)ノ゙】BZOJ1531-[POI2005]Bank notes
    【可持久化并查集】BZOJ3673-可持久化并查集 by zky
    【manacher+FFT】BZOJ3160-万径人踪灭
    SCOI2008着色方案(记忆化搜索)
    [JSOI2008]魔兽地图(树形dp)
    NOI2017蔬菜(贪心)
  • 原文地址:https://www.cnblogs.com/kuangbin/p/3315075.html
Copyright © 2011-2022 走看看