3 条题解

  • 1
    @ 2025-4-20 12:10:44

    最难的代码 最简洁的代码

    #include<iostream>
    using namespace std;
    int main ()
    {
    	int a;
    	cin >> a;
    	if(a>=90)
    	{
    		cout << "A";
    	}
    	
    	if(a>=80&a<90)
    	{
    		cout << "B";
    	}
    	if(a>=70&a<80)
    	{
    		cout << "C";
    	 } 
    	 if(a>=60&a<70)
    	{
    		cout << "D";
    	}
    	if(a<60)
    	{
    		cout << "E";
    	}
    }
    
    
    • 1
      @ 2025-3-11 13:26:46
      #include <queue>
      #include <math.h>
      #include <stack>
      #include <stdio.h>
      #include <iostream>
      #include <vector>
      #include <iomanip>
      #include <string.h>
      #include <algorithm>
      using namespace std;
      #define elif else if
      #define int long long
      #define float double
      #define N 30000010
      #define INF 0xc00000fd
      #define scf scanf
      #define ptf printf
      #define gtc getchar
      #define ptc putchar
      #define psp ptc(' ')
      #define pel ptc('\n')
      #define pzr ptc('0')
      #define fin for(int i = 1; i <= n; i++)
      #define fim for(int i = 1; i <= m; i++)
      #define fjn for(int j = 1; j <= n; j++)
      #define fjm for(int j = 1; j <= m; j++)
      #define rai fin	a[i] = read()
      #define wct write(cnt)
      #define pct print(cnt)
      #define was write(ans)
      #define pas print(ans)
      queue<int> Q;
      stack<int> S;
      vector<int> V;
      inline int wei(int n){
      	int cnt = 0;
      	while(n > 0){
      		n /= 10;
      		cnt++;
      	}
      	return cnt;
      }
      inline bool isPrime(int n){
      	if(n < 2)
      		return 0;
      	for(int i = 2; i * i <= n; ++i)
      		if(n % i == 0)
      			return 0;
      	return 1;
      }
      inline int read(string n){
      	int x = 0, f = 1;
      	ptf("%s", n.c_str());
      	char c = gtc();
      	while(c < '0'  ||  c > '9'){
      		if(c == '-')
      			f = -1;
      		c = gtc();
      	}
      	while(c >= '0'  &&  c <= '9'){
      		x = x * 10 + c - 48;
      		c = gtc();
      	}
      	return x * f;
      }
      inline float input(string n){
      	float x = 0, f = 1, x2 = 0, cnt = 0, i = 0;
      	ptf("%s", n.c_str());
      	char c = gtc();
      	while(c < '0'  ||  c > '9'){
      		if(c == '-')
      			f = -1;
      		c = gtc();
      	}
      	while(c >= '0'  &&  c <= '9'){
      		x = x * 10 + c - 48;
      		c = gtc();
      	}
      	c = gtc();
      	while(c >= '0'  &&  c <= '9'){
      		x2 = x2 * 10 + c - 48;
      		cnt++;
      		c = gtc();
      	}
      	for(; i < cnt; i++)
      		x2 /= 10.0;
      	return (x + x2) * f;
      }
      inline void write(int n) {
          if(n < 0){
              ptc('-');
              n = -n;
          }
          if(n > 9)
      		write(n / 10);
          ptc(n % 10 + '0');
      	return;
      }
      inline void print(int n, float m){
      	write(m);
      	m -= (int)m;
      	ptc('.');
      	int cnt = 0;
      	while(m - (int)m && n - cnt){
      		m *= 10;
      		ptc((int)m % 10 + '0');
      		cnt++;
      	}
      	n -= cnt;
      	fin
      		ptc('0');
      	return;
      }
      #define input() input("")
      #define read() read("")
      int n = read();
      char c;
      inline void Main(){
      	if(n > 90)
      		c = 'A';
      	elif(n >= 80)
      		c = 'B';
      	elif(n >= 70)
      		c = 'C';
      	elif(n >= 60)
      		c = 'D';
      	else
      		c = 'E';
      	ptc(c);
      	return;
      }
      signed main(signed argc, char **argv){
      	Main();
      	pel;
      	return 0;
      }
      
      • -2
        @ 2022-1-21 23:27:19
        #include <iostream>
        using namespace std;
        int main(){
            int n;
            cin>>n;
            if(n>=90)
            cout<<"A";
            else if(n>=80)
                cout<<"B";
            else if(n>=70)
                cout<<"C";
            else if(n>=60)
                cout<<"D";
            else    
                cout<<"E";
            return 0;
        }
        • 1

        信息

        ID
        854
        时间
        1000ms
        内存
        128MiB
        难度
        2
        标签
        递交数
        68
        已通过
        44
        上传者