微软近日推出了一款功能极简的手机,在手机上用一个包含了 7 \times 77×7 个像素的区域来显示手机信号。满信号的时候显示如下:
1 +-----+ 2 |- 4G| 3 |-- | 4 |--- | 5 |---- | 6 |-----| 7 +-----+每一格信号(第 i(1 \le i \le 5)i(1≤i≤5) 格信号有 ii 个-)代表 20\%20% 的信号强度,不足一格信号的部分不显示。同时会在右上角显示当前的网络传输模式。在信号强度不低于 90\%90% 的时候显示4G;当信号低于 90\%90%、不低于 60\%60% 的时候显示3G;否则显示E。
对于给定的当前信号强度 d\%d%,输出信号的 7 \times 77×7 像素的图案。
输入一个整数 d(0 \le d \le 100)d(0≤d≤100),表示信号强度。
按照题目要求输出,每行末尾不要输出多余的空白字符。
题目大意:中文题
解题思路:模拟
#include<iostream> #include<cstring> #include<string> #include<vector> #include<algorithm> #include<cstdio> #include<map> #include<set> #include<cmath> #include<cctype> #include<cstdlib> #include<list> #include<iomanip> using namespace std; typedef long long LL; const int MAXN=1e3+10; const int INF=0x3f3f3f3f; int main() { int d; while(cin>>d) { if(d<20) { cout<<"+-----+"<<endl; cout<<"| E|"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"+-----+"<<endl; }else if(d>=20&&d<40) { cout<<"+-----+"<<endl; cout<<"|- E|"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"+-----+"<<endl; }else if(d>=40&&d<60) { cout<<"+-----+"<<endl; cout<<"|- E|"<<endl; cout<<"|-- |"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"+-----+"<<endl; }else if(d>=60&&d<80) { cout<<"+-----+"<<endl; cout<<"|- 3G|"<<endl; cout<<"|-- |"<<endl; cout<<"|--- |"<<endl; cout<<"| |"<<endl; cout<<"| |"<<endl; cout<<"+-----+"<<endl; }else if(d>=80&&d<90) { cout<<"+-----+"<<endl; cout<<"|- 3G|"<<endl; cout<<"|-- |"<<endl; cout<<"|--- |"<<endl; cout<<"|---- |"<<endl; cout<<"| |"<<endl; cout<<"+-----+"<<endl; }else if(d>=90&&d<100) { cout<<"+-----+"<<endl; cout<<"|- 4G|"<<endl; cout<<"|-- |"<<endl; cout<<"|--- |"<<endl; cout<<"|---- |"<<endl; cout<<"| |"<<endl; cout<<"+-----+"<<endl; }else { cout<<"+-----+"<<endl; cout<<"|- 4G|"<<endl; cout<<"|-- |"<<endl; cout<<"|--- |"<<endl; cout<<"|---- |"<<endl; cout<<"|-----|"<<endl; cout<<"+-----+"<<endl; } } return 0; }