Question

这是在网上法官上提交的一部法典,用于生成主要数字,但Im正在陷入分化错误。目标是generate prime number between the given range m to n (with n > m)。采用Etosthenes算法实施。请告诉我,什么情况会错。感谢:

#include <stdio.h>
#include <math.h>

int main(){
    long int m,n,c1,c2,c3;
    int t;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&m,&n);


        //create prime list
        short int *prime;
        prime = (short int*)malloc((n-m)*sizeof(short int));

        //fill list with 0 - prime
        for(c1 = 2; c1 <= n; c1++){
                prime[c1] = 1;
        }

        //set 1 and 0 as not prime
        prime[0]=0;
        prime[1]=0;

        //find primes then eliminate their multiples (0 = prime, 1 = composite)
        for(c2 = 2;c2 <= (int)sqrt(n)+1;c2++){
            if(prime[c2]){
                c1=c2;
                for(c3 = 2*c1;c3 <= n; c3 = c3+c1){
                    prime[c3] = 0;
                }
            }
        }

        //print primes
        for(c1 = m; c1 <=n; c1++){
            if(prime[c1]) printf("%d
",c1);
        }
    }       
    return 0;
}

Answer 1

c3 can go up to n in the innermost loop, but you only may allocate less than n slots in your array. In fact, even if you allocated n slots, index n would be one more than the number of slots you allocated. At worst, you d just corrupt some memory past the end of the array and hopefully not trash the stack. At best, I guess you get a segfault. You probably want to change your X <= n to X < n or allocate one more element in your array. In fact, you probably should just allocate (n + 1) * sizeof(short) bytes for your array.

而且,你从未放过头,你从来从未验证用户的投入。如果是竞争,而竞争会限制投入,则后者可能oka。另外,你从未放过<代码>prime阵列,因此你有传闻。

Answer 2

Of course it will you are allocating a memory which is (n-m) & you are acessing prime[n] ,

Answer 3

如果只有1个要素的时间长,可以避免:

//set 1 and 0 as not prime
        prime[0]=0;
        prime[1]=0;

Answer 4

You malloc(n-m),但在以下地方,你开始做主[2.n]。 n-m在1E5中,但n本身可以是1E9(相当数量)。你的简单想法可能无法落实,因为小鼠(1E9)可能失败。你们需要一种易法算法。

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