zoukankan      html  css  js  c++  java
  • xtu read problem training 3 A

    The Child and Homework

    Time Limit: 1000ms
    Memory Limit: 262144KB
    This problem will be judged on CodeForces. Original ID: 437A
    64-bit integer IO format: %I64d      Java class name: (Any)
     
     
     
    Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.

    Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:

    • If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
    • If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).

    You are given a multiple-choice questions, can you predict child's choose?

    Input

    The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.

    Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".

     

    Output

    Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).

     

    Sample Input

    Input
    A.VFleaKing_is_the_author_of_this_problem
    B.Picks_is_the_author_of_this_problem
    C.Picking_is_the_author_of_this_problem
    D.Ftiasch_is_cute
    Output
    D
    Input
    A.ab
    B.abcde
    C.ab
    D.abc
    Output
    C
    Input
    A.c
    B.cc
    C.c
    D.c
    Output
    B

    Hint

    In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.

    In the second sample, no choice is great, so the child will choose the luckiest choice C.

    In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B.

     

    Source

     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <climits>
     7 #include <vector>
     8 #include <queue>
     9 #include <cstdlib>
    10 #include <string>
    11 #include <set>
    12 #define LL long long
    13 #define INF 0x3f3f3f3f
    14 using namespace std;
    15 struct node {
    16     char str[1000];
    17     int len,index;
    18 };
    19 bool cmp(const node &x,const node &y) {
    20     return x.len < y.len;
    21 }
    22 node p[4];
    23 int main() {
    24     int i,j;
    25     for(i = 0; i < 4; i++) {
    26         scanf("%s",p[i].str);
    27         p[i].len = strlen(p[i].str)-2;
    28         p[i].index = i;
    29     }
    30     sort(p,p+4,cmp);
    31     for(i = 1; i < 4; i++) {
    32         if(p[i].len < p[0].len*2) break;
    33     }
    34     for(j = 0; j < 3; j++)
    35         if(p[j].len*2 > p[3].len) break;
    36     if(i == 4 && j == 3 || i < 4 && j < 3) puts("C");
    37     else printf("%c
    ",i == 4?p[0].index+'A':p[3].index+'A');
    38     return 0;
    39 }
    View Code
  • 相关阅读:
    《设计模式》-原则二:里氏代换原则(LSP)
    设计模式从0开始
    net reactor加密源码保软件安全-net reactor使用教程
    python中import和from...import区别
    Python之import
    c#执行bat批处理文件,并通过线程将结果显示在控件中
    C#中双问号、双冒号等几个特殊关键字
    Eclipse debug高级技巧(转)
    Android从零开始--安装
    以另一个用户来运行程序
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/3891599.html
Copyright © 2011-2022 走看看