항 저 우 전기 ACM 1000, 1001, 1002 자바 해답
4936 단어 Acm
Input Each line will contain two integers A and B. Process to end of file.
Output For each case, output A + B in one line.
파일 의 끝 에 프로 세 스 를 주의 하 십시오.
import java.util.Scanner;
public class Main {
public static void main(String args[]){
int a,b;
Scanner s=new Scanner(System.in);
while(s.hasNextInt()){
a=s.nextInt();
b=s.nextInt();
System.out.println((a+b));
}
}
}
1001 . Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).
In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + … + n.
Input The input will consist of a series of integers n, one integer per line.
Output For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.
n * (n - 1) 을 계산 하면 넘 칠 수 있 으 니 주의 하 세 요.
import java.util.Scanner;
public class Main {
public static void main(String args[]){
Scanner s=new Scanner(System.in);
while(s.hasNextInt()){
int n=s.nextInt();
if((n<<31)>>31==0){
int sum=n/2;
sum=sum*(n+1);
System.out.println(sum);
System.out.println();
}
else{
int sum=(n+1)/2;
sum=sum*n;
System.out.println(sum);
System.out.println();
}
}
}
}
1002.
Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
마지막 출력 을 주의 한 후 줄 을 바 꿀 필요 가 없습니다.
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int T=s.nextInt();
for(int i=0;iout.println("Case "+(i+1)+":");
System.out.println(a.toString()+" + "+b.toString()+" = "+c.toString());
if(i!=T-1)
System.out.println();
}
}
}