for语句计算1~100之间所有连续整数的和

var num = 1; var num1 = 1; var oddSum = 0; var evenSum = 0; var timesSum = 0; while(num <= 100){ if (num % 2 == 0) { evenSum = evenSum + num; }else if (num % 2 == 1) { oddSum = oddSum + num; } num++; } while (num1 <= 100){ if (num1 % 3 == 0) { times

#include<stdio.h> int main(agrc *agrv) { int n,i; int sum=0; scanf("%d",&n); for(i=1;i<=n;i++){ if(i%3==0&&i%7!=0){ sum+=i; } } printf("%d\n",sum); } 原文地址:https://www.cnblogs.com/haidaxiaobaobao/p/10040039.html

//1.建立一个for循环用于输入数据,设置退出条件 //2.算出平均成绩 #include<iostream> using namespace std; int main() { int Score,sum=0,k=0; int a[100]; float Average; cout<<"please input some students's score:"<<endl; for(int i=0;i<100;i++) { cin>&g

计算1到100之间和奇数和与偶数和 #!/bin/bash Evensum=0 ODDsum=0 for i in {1..100};do if [ $(( i%2 )) -eq 0 ];then let Evensum+=$i else let ODDsum+=$i fi done echo "Evensum is $Evensum" echo "ODDsum is $ODDsum" ~ 原文地址:https://blog.51cto.com/11342825/2

//利用switch语句计算特定的年份的月份共有几天. let year =2015 let month =2 //先判断闰年中二月份的情况 ifmonth ==2 { if (year %400 ==0||year %4 ==0 &&year %100 !=0) //闰年的二月份29天,非闰年的二月份是28天 { print("29天") }else { print("28天") } } else { //考虑其他两种情况 switchmonth

方法一:利用一个布尔类型的值的判断,优点想法更直接易懂 public static void main(String[] args) { for(int i=2;i<=100;i++){ boolean b=true; for(int n=2;n<i;n++){ if(i%n==0){ b=false; break; } } if(b==true){ System.out.print(i+"\t"); } } } 方法二:能被2和这个数本身以外的数整除的都不是素数.理解难但

1.计算1~100的数之和----for循环实现1~100的和 sum1 = 0 for i in range(1,101): sum1 = sum1 + i i += 1 print(f"1-100之间的和是:{sum1}") ----while实现1~100的和 sum1 = 0 i = 1 while True: sum1 = sum1 + i if i == 100: break i += 1 print(sum1) 2.计算1~100里面偶数之和 sum1 = 0 for i

1 /* 2 * 求1到100之间的质数 3 */ 4 public class Zy3 { 5 public static void main(String[] args) { 6 System.out.print("1到100之间的素数有:\n"+"\t"+1+"\n"); 7 int j=0; 8 for(int i=2;i<100;++i){ 9 for(j=2;j<i;++j){ 10 if(i%j==0) 11 break

#include <cstdio> int main() { int i = 0, j = 0; while (i <= 100 && j <= 100) { if (i < j) { printf("(%d) ", i); i += 3; } else if (i > j) { printf("[%d] ", j); j += 5; } else { printf("{%d} ", i); i