This question requires the realization of a simple function for judging prime numbers and a function for calculating the sum of prime numbers in a given interval using this function.
A prime number is a positive integer that can only be divisible by 1 and itself. Note: 1 is not a prime number, 2 is a prime number.
Function interface definition:
int prime( int p );
int PrimeSum( int m, int n );
The function prime returns 1 when the user passes in the parameter p as a prime number, otherwise it returns 0; the function PrimeSum returns the sum of all prime numbers in the interval [m, n]. The title guarantees that the parameter m ≤ n passed in by the user.
Sample referee test procedure:
#include <stdio.h>
#include <math.h>
int prime( int p );
int PrimeSum( int m, int n );
int main()
{
int m, n, p;
scanf("%d %d", &m, &n);
printf("Sum of ( ");
for( p=m; p<=n; p++ ) {
if( prime(p) != 0 )
printf("%d ", p);
}
printf(") = %d\n", PrimeSum(m, n));
return 0;
}
/* 你的代码将被嵌在这里 */
Input sample:
1 10
Sample output:
Sum of ( 2 3 5 7 ) = 17
answer:
int prime( int p )
{
//参数p为素数时返回1，否则返回0
int i;
int t = 1;
for(i = 2 ; i <= sqrt(p) ; i++) //详细说明看下个函数
{
if(p % i == 0)
{
t = 0;
break;
}
}
if(t == 1 && p != 1 && p>0) return 1;
else return 0;
}
int PrimeSum( int m, int n )
{
//返回区间[m, n]内所有素数的和
int i,s; //循环变量
int t=1; //中介指针，如果t=1表示是素数，t=0就i不是素数
int sum=0; //素数总和
for(i=m;i<=n;i++) //区间内循环
{
t=1; //没判断完一次素数，中介重新赋值为1
for(s=2;s<=sqrt(i);s++) //每一个素数是从2开始到它的平方根
{
if(i%s==0) //能被其他数整除就不是素数了
{
t=0; //t变为0表示不是素数
break; //退出内循环
}
}
if(t==1 && i!=1 && i>0) sum+=i; //这里素数的条件是：t等于1 并且 该数不能是1(1不是素数) 并且 该数要大于0(因为我们输入的数可以小于0)
}
return sum;
}