PAT基础编程题目-7-14 求整数段和

题目详情

题目地址：https://pintia.cn/problem-sets/14/problems/794

解答

C语言版

#include<stdio.h>
int main() {
   
	int a, b, sum = 0, count=1;
	scanf("%d %d", &a, &b);
	for (int i = a; i <= b; i++, count++)
	{
   
		printf("%5d", i);
		if (count % 5 == 0)
			printf("\n");
		sum = sum + i;
	}
	if((count-1)%5==0)  // 避免多换行一次
		printf("Sum = %d", sum);
	else 
	    printf("\nSum = %d", sum);
	return 0;
}

C++版

#include<iostream>
#include<iomanip>
using namespace std;
int main() {
   
	int a, b, sum = 0, count = 1;
	cin >> a >> b;
	for (int i = a; i <= b; i++, count++)
	{
   
		cout << setw(5) << i;  //设置宽度setw
		if (count % 5 == 0)
			cout << endl;
		sum = sum + i;
	}
	if ((count - 1) % 5 == 0)  // 避免多换行一次
		cout << "Sum = " << sum;
	else
		cout << endl << "Sum = " << sum;
	return 0;
}

Java版

import java.util.Scanner;
public class Main{
   

	public static void main(String[] args) {
   
		int a = 0, b = 0, sum = 0, count = 1;
		Scanner scanner = new Scanner(System.in);
		if (scanner.hasNext()) {
   
			a = scanner.nextInt();
			b = scanner.nextInt();
		}
		scanner.close();
		for (int i = a; i <= b; i++, count++)
		{
   
			/** * String.format()： * 可对整数进行格式化：%[index$][标识][最小宽度]转换方式 * %[index$]：以%index$开头，index从1开始取值，表示将第index个参数拿进来进行格式化 * [标识]：' ' 正值前加空格，负值前加负号 * [最小宽度]：最终该整数转化的字符串最少包含多少位数字 * 转换方式：d-十进制 o-八进制 x或X-十六进制 */
			System.out.print(String.format("%1$ 5d", i));  
			if (count % 5 == 0)
				System.out.println();
			sum = sum + i;
		}
		if ((count - 1) % 5 == 0)  // 避免多换行一次
			System.out.print("Sum = "+sum);
		else
			System.out.print("\nSum = "+sum);;

	}

}

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