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<>习题5-4 使用函数求素数和 (20 分)
本题要求实现一个判断素数的简单函数、以及利用该函数计算给定区间内素数和的函数。
素数就是只能被1和自身整除的正整数。注意：1不是素数，2是素数。
<>函数接口定义：
int prime( int p );
int PrimeSum( int m, int n );
其中函数prime当用户传入参数p为素数时返回1，否则返回0；函数PrimeSum返回区间[m, n]内所有素数的和。题目保证用户传入的参数m≤n。
<>裁判测试程序样例：
#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 */
<>输入样例：
-1 10
<>输出样例：
Sum of ( 2 3 5 7 ) = 17
<>思路：
大于1，且只能被1和自己除的数为素数
#include <stdio.h> #include <math.h> int prime( int p )//素数返回1 { int i,flag=0;
for(i=2;i<p;i++) { if(p%i==0) break; } if(i==p) flag=1; return flag; } int
PrimeSum( int m, int n ) { int i,sum=0; for(i=m;i<=n;i++) { if(prime(i)==1) sum=
sum+i; } return sum; } 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; }
运行结果：
