("%d", &numbers[i]); sum += numbers[i]; // 累加和 // 更新最大值索引 if (numbers[i] > numbers[max_index]) { max_index = i; } } // 计算平均数 average = sum / 10.0; // 输出结果 printf("平均数为: %.2f\n", average); printf("最大数为: %d\n", numbers[max_index])...
Program to reverse a String Number Crunching Program to find Average of n Numbers Armstrong Number Checking input number for Odd or Even Print Factors of a Number Find sum of n Numbers Print first n Prime Numbers Find Largest among n Numbers Exponential without pow() method Find whether numbe...
intmain(){ intn,sum=0; printf("This program sums a series of integers.\n"); printf("Enter integers (0 to terminate):"); scanf("%d",&n); while(n!=0){ sum+=n; scanf("%d",&n); } printf("The sum is: %d\n",sum); return0; } 6.2 语句 do语句和while语句关系紧密。事实上,...
C - Swap two numbers W/O using a temporary variable using C program? C - Read name & marital status of a girl & print her name with Miss or Mrs C - Check given number is divisible by A & B C - Find sum of all numbers from 0 to N W/O using loop C - Input hexadecimal valu...
printf("Sum: %d",sum); /* Displays sum */ return 0; } 输出 Enter two integers: 12 11 Sum: 23 4、C语言实现两个小数相乘 源代码: /*C program to multiply and display the product of two floating point numbers entered by user. */ ...
printf("Address of Parameter: %p\n", ¶m); return 0; } Parameter Value: 5.000010000200003190684583387338 Address of Parameter: 0x7fffffffddf0 [wenxue@hpi7 hellvsc]$ /// #include <stdio.h> int main() { long double* ptr_ld_var, ld_var; ld_var = 5.00001000020000300004000050000600007; /...
原文:https://beginnersbook.com/2014/06/c-program-to-check-armstrong-number/ 如果数字的各位的立方和等于数字本身,则将数字称为阿姆斯特朗数。在下面的 C 程序中,我们检查输入的数字是否是阿姆斯特朗数。 #include<stdio.h>intmain(){intnum,copy_of_num,sum=0,rem;//Store input number in variable numpri...
In this article, we will learn how to write a C program to find sum of array elements. For example, if the array is [1, 2, 3, 4, 5] then the program should print 1+2+3+4+5 = 15 as output. We will see various different ways to accomplish this. Example 1:
if (numbers[j] % 4 == 0) { sum = sum + numbers[j]; } } printf("数组中能被4整除的数的累加和是:%d\n", sum); return 0; } 此代码说明。 1. 数组定义:定义了一个整数数组 `numbers` 并初始化了一些值。 2. 计算数组大小:通过 `sizeof(numbers) / sizeof(numbers[0])` 计算数组元素...
inta[10000],b[10000],i,n,c; printf("Enter size of the array : "); scanf("%d",&n); printf("Enter elements in array : "); for(i=0;i<n;i++) { scanf("%d",&a[i]); } c=count(a,n); printf("duplicate numbers in the array: %d",c); ...