题目一：2000.ASCLL码排序
Problem Description
输入三个字符后，按各字符的ASCII码从小到大的顺序输出这三个字符。
Input
输入数据有多组，每组占一行，有三个字符组成，之间无空格。
Output
对于每组输入数据，输出一行，字符中间用一个空格分开。
#include<iostream> using namespace std; int main() { char a, b, c; while(cin >> a >> b >> c) { if(a > b) { swap(a, b); // 如果a的ASCII值大于b，交换a和b } if(a > c) { swap(a, c); // 如果交换后a的ASCII值仍然大于c，交换a和c } if(b > c) { swap(b, c); // 确保b的ASCII值不大于c } cout << a << " " << b << " " << c << endl; } return 0; }
题目二：2051.Bitset
Problem Description
Give you a number on base ten,you should output it on base two.(0 < n < 1000)
Input
For each case there is a postive number n on base ten, end of file.
Output
For each case output a number on base two.
#include <iostream> #include <string> using namespace std; int main() { int n; while(cin >> n) { if (n < 1 || n > 1000) { return 1; } string s = ""; while (n > 0) { s= to_string(n % 2) + s; n /= 2; } cout << s<< endl; } return 0; }
题目三：2052.Picture
Problem Description
Give you the width and height of the rectangle,darw it.
Input
Input contains a number of test cases.For each case ,there are two numbers n and m (0 < n,m < 75)indicate the width and height of the rectangle.Iuput ends of EOF.
Output
For each case,you should draw a rectangle with the width and height giving in the input.
after each case, you should a blank line.
#include <iostream> #include <iomanip> using namespace std; int main() { int w, h; while (cin >> w >> h) { if (w <= 0 || h <= 0 || w > 75 || h > 75) { continue; } cout << '+'; for (int i = 0; i < w ; ++i) { cout << '-'; } cout << '+' << endl; for (int i = 0; i < h ; ++i) { cout << '|'; for (int j = 0; j < w; ++j) { if (j == 0 || j <= w ) { cout << ' '; } else { cout << '|'; } } cout << '|' << endl; } cout << '+'; for (int i = 0; i < w ; ++i) { cout << '-'; } cout << '+' << endl; cout << endl; } return 0; }
题目四：2053.Switch Game
Problem Description
There are many lamps in a line. All of them are off at first. A series of operations are carried out on these lamps. On the i-th operation, the lamps whose numbers are the multiple of i change the condition ( on to off and off to on ).
Input
Each test case contains only a number n ( 0< n<= 10^5) in a line.
Output
Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).
#include <iostream> #include <cmath> using namespace std; int main() { int n; while (cin >> n) { int sqrt_n = sqrt(n); // 由于可能存在精度问题，我们需要检查sqrt_n的平方是否等于n if (sqrt_n * sqrt_n == n) { // 如果是完全平方数，则第n个灯最终是开的 cout << 1 << endl; } else { // 否则，第n个灯最终是关的 cout << 0 << endl; } } return 0; }
题目五：2054.A= =B？
Problem Description
Give you two numbers A and B, if A is equal to B, you should print "YES", or print "NO".
Input
each test case contains two numbers A and B.
Output
for each case, if A is equal to B, you should print "YES", or print "NO".
#include<iostream> #include<string> using namespace std; void senre(string &a){ int temp=0; if(a.find('.')!=a.npos){ while(*(a.end()-1-temp)=='0'){ temp++; } a.resize(a.size()-temp); if(*(a.end()-1)=='.'){ a.resize(a.size()-1); } } while(*(a.begin())=='0'){ a.erase(a.begin()); } } int main() { string a; string b; while(cin>>a>>b){ senre(a); senre(b); if(a==b){ cout<<"YES"<<endl; } else{ cout<<"NO"<<endl; } } return 0; }
这个题目不知道为什么结果一直不对， 后了解应是题目中对数据的限制比较小，要判断的数据可能是小数，精度问题等，应该转化为字符串进行比较，验证了我们要时刻注意着数据的边界条件以及题干中的限制范围，合理输出。
后面修改了这个题目代码，过了
