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Ningbo [1219] Time(将数字转换成时钟那样的数字)
阅读量:4036 次
发布时间:2019-05-24

本文共 3500 字,大约阅读时间需要 11 分钟。

1、

2、比赛的时候居然想了一个小时,才意识到是3*3字符的含义,纠结,好在一遍AC了

题目:

  • [1219] Time

  • 时间限制: 1000 ms 内存限制: 131072 K
  • 问题描述
  • Digital clock use 4 digits to express time, each digit is described by 3*3 characters (including”|”,”_”and” “).now given the current time, please tell us how can it be expressed by the digital clock.
  • 输入
  • There are several test cases.
    Each case contains 4 integers in a line, separated by space.
    Proceed to the end of file.
  • 输出
  • For each test case, output the time expressed by the digital clock such as Sample Output.
  • 样例输入
  • 1 2 5 62 3 4 2
  • 样例输出
  • _  _  _   | _||_ |_   ||_  _||_| _  _     _  _| _||_| _||_  _|  ||_
  • 提示
  • The digits showed by the digital clock are as follows:   _  _     _  _  _  _  _  _  | _| _||_||_ |_   ||_||_|| | ||_  _|  | _||_|  ||_| _||_|
  • 来源
  • 辽宁省赛2010
3、AC代码:

#include
#include
#include
#include
#include
using namespace std;int a[5];int main(){ while(scanf("%d%d%d%d",&a[0],&a[1],&a[2],&a[3])!=EOF) { for(int i=0; i<3; i++) { for(int j=0; j<=3; j++) { if(i==0) { if(a[j]==1) printf(" "); else if(a[j]==2) printf(" _ "); else if(a[j]==3) printf(" _ "); else if(a[j]==4) printf(" "); else if(a[j]==5) printf(" _ "); else if(a[j]==6) printf(" _ "); else if(a[j]==7) printf(" _ "); else if(a[j]==8) printf(" _ "); else if(a[j]==9) printf(" _ "); else if(a[j]==0) printf(" _ "); } else if(i==1) { if(a[j]==1) printf(" |"); else if(a[j]==2) printf(" _|"); else if(a[j]==3) printf(" _|"); else if(a[j]==4) printf("|_|"); else if(a[j]==5) printf("|_ "); else if(a[j]==6) printf("|_ "); else if(a[j]==7) printf(" |"); else if(a[j]==8) printf("|_|"); else if(a[j]==9) printf("|_|"); else if(a[j]==0) printf("| |"); } else if(i==2) { if(a[j]==1) printf(" |"); else if(a[j]==2) printf("|_ "); else if(a[j]==3) printf(" _|"); else if(a[j]==4) printf(" |"); else if(a[j]==5) printf(" _|"); else if(a[j]==6) printf("|_|"); else if(a[j]==7) printf(" |"); else if(a[j]==8) printf("|_|"); else if(a[j]==9) printf(" _|"); else if(a[j]==0) printf("|_|"); } } printf("\n"); } } return 0;}

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