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解题思路:
- int prime(int n)
- { int k,flag=1;
- for (k=2; k<=(int)sqrt((double)n); k++)
- if (n%k == 0)
- flag=0;
- return flag;
- }
- int fun(int m, int a[])
- {
- int k, s, count, i=0;
- for(k=6; k<=m; k+=2)
- { count = 0;
- /* 请在此处填写代码 */
- for(s=2;s<=k/2;s++)
- { if(prime(s)&&prime(k-s))
- {
- count++;}}
- if (count == 10) {
- printf("%d\n", k);
- a[i++] = k;
- }
- }
- return i;
- }

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