개발일지📝

#include using namespace std; int main(){ int a; scanf("%d", &a); if (a=90) printf("A"); else if (a=80) printf("B"); else if (a=70) printf("C"); else if (a=60) printf("D"); else printf("F"); return 0; }

#include using namespace std; int main(){ int a,b; cin >> a >> b ; if (a>b) cout

import java.util.Scanner; public class Main { public static void main(String[] args){ Scanner sc = new Scanner(System.in); int a = sc.nextInt(); int b = sc.nextInt(); //45 + if (b >= 45) { b-=45; System.out.printf("%d %d", a ,b); } else if (a == 0) { a += 23; b = (60+b)-45; System.out.printf("%d %d", a ,b); } else { a -= 1; b = (60+b)-45; System.out.printf("%d %d", a ,b); } } } 조건만 잘 생각해서 코딩해주면 ..

import java.util.Scanner; public class Main { public static void main(String[] args){ Scanner sc = new Scanner(System.in); int a = sc.nextInt(); int b = sc.nextInt(); //++ 1 if ( a >0 && b >0) System.out.printf("1"); //-+ 2 else if ( a 0) System.out.printf("2"); // -- 3 else if ( a 0 && b

import java.util.Scanner; public class Main { public static void main(String[] args){ Scanner sc = new Scanner(System.in); int a = sc.nextInt(); if (a%4 == 0){ //4의배수이고 if (a%100 != 0){ System.out.printf("1"); } else if (a%400 == 0){ System.out.printf("1"); } else { System.out.printf("0"); } } else System.out.printf("0"); } }​ import java.util.Scanner; public class Main { public static void main..

import java.util.Scanner; public class Main { public static void main(String[] args){ Scanner sc = new Scanner(System.in); int a = sc.nextInt(); if (a=90) System.out.printf("A"); else if (a=80) System.out.printf("B"); else if (a=70) System.out.printf("C"); else if (a=60) System.out.printf("D"); else System.out.printf("F"); } }

import java.util.Scanner; public class Main { public static void main(String[] args){ Scanner sc = new Scanner(System.in); int a = sc.nextInt(); int b = sc.nextInt(); if (a>b) { System.out.printf(">"); } else if (a

#include using namespace std; int main(){ int A,B; scanf("%d %d ", &A, &B); printf("%d\n",A*(B%10)); printf("%d\n",A*((B/10)%10)); printf("%d\n",A*(B/100)); printf("%d",A*B); return 0; }

#include #include using namespace std; int main(){ int a,b; scanf("%d %d", &a,&b); printf("%d\n",a+b); printf("%d\n",a-b); printf("%d\n",a*b); printf("%d\n",a/b); printf("%d",a%b); return 0; }

#include #include using namespace std; int main(){ double a,b; scanf("%lf %lf", &a, &b); printf("%.10lf",a/b); return 0; } 상대오차가 10-9 이하이면 정답이다. => 출력범위 조절