{ sum+=data[i];printf("%d",data[i]);} } printf("s=%d\n",sum);return0;}
结果1 题目The sum of two numbers is 10. One number is 4. What is the other number? A. 4 B. 5 C. 6 D. 7 相关知识点: 试题来源: 解析 C。两数之和是 10,一个数是 4,另一个数是 10 减 4 等于 6,所以选 C。反馈 收藏
The sum of two numbers is 15. One number is 8. What is the other number? A. 5 B. 6 C. 7 D. 8 相关知识点: 试题来源: 解析 C。选项 A,5 是错误的,因为 8 加 5 不等于 15。选项 B,6 是错误的,8 加 6 也不等于 15。选项 C,7 是正确的,8 加 7 等于 15。选项 D,8 是错误...
Two Sum - Difference equals to target Given an array of integers, find two numbers that their difference equals to a target value. where index1 must be less than index2. Please note that your returned answers (both index1 and in......
This C language program collection has more than 100 programs, covering beginner level programs like Hello World, Sum of Two numbers, etc. to complex programs like Fibonacci series, Prime Numbers, and pattern printing programs. All the programs have working code along with their output. The ...
The sum of any two numbers is the addition of the value of the two given numbers. In C language, the addition of any number is done by using the arithmetic operator. This operation can be performed on various data types like integer, float, double, etc....
解析 C For example:2、3、5、7 are all primes,2+3=5,2+5=7,5 and 7 are prime and odd number;3+5=8,3+7=10,8和10 are composite number and even number.So the sum of two prime numbers will be an odd or even number.So we choose ....
百度试题 结果1 题目The sum of two numbers is 15. One number is 7. What is the other number? A. 8 B. 7 C. 6 D. 9 相关知识点: 试题来源: 解析 A。两个数的和是 15,一个数是 7,另一个数就是 15 减去 7 等于 8 。反馈 收藏 ...
printf("%d can't be expressed as sum of two prime numbers.",n); return 0; } int prime(int n) /* Function to check prime number */ { int i, flag=1; for(i=2; i<=n/2; ++i) if(n%i==0) flag=0; return flag; }
("Enter 7 numbers:\n"); for(i = 0; i < 7; ++i) { //try scanf("%d", &x[i]); scanf("%lf", x+i); // try sum += x[i] sum += *(x+i); } printf("Sum = %lf\n", sum); return 0; } Enter 7 numbers: 1.1 2.2 3.3 4.4 5.5 6.6 7.7 Sum = 30.800000 [wenxue@...