给定一个元素数组,任务是打印那些数字总和也是质数且返回-1的数字,数组中不存在这样的数字
Input: arr[]={2,4,3,19,25,6,11,12,18,7} Output : 2, 3, 25, 11, 12, 7
在这里,给定的输出将生成,因为它包含了它们的和也为质数的可加数,例如− 2、3、7是质数,但25(2 + 5 = 7),11(1 + 1 = 2),12(1+ 2 = 3)也是素数,而像19(1 + 9 = 10)这样的数字不是素数。
START Step 1 -> Take array of int with values Step 2 -> declare start variables as i, m, flag, flag1, sum, r, d, j, tem Step 3 -> store size of array in m as sizeof(arr)/sizeof(arr[0]) Step 4 -> Loop For i=1 and i<m and i++ Set flag=flag1=sum=0 Set d=int(arr[i]/2 Loop For j=2 and j<=d and j++ IF arr[i]%j==0 Set flag=1 Break End IF End IF flag=0 Set tem=arr[i] Loop While tem Set r=tem%10 Set sum=sum+r Set tem=tem/10 End Set d=int(sum/2) Loop For j=2 and j<=d and j++ IF sum%j=0 Set flag1=1 break End End IF flag1=0 Print arr[i] End End End STOP
#include<iostream> using namespace std; int main(){ int arr[]={2,4,3,19,25,6,11,12,18,7}; int i,m,flag,flag1,sum,r,d,j,tem; m=sizeof(arr)/sizeof(arr[0]); for(i=0;i<m;i++) { flag=flag1=sum=0; d=int(arr[i]/2); for(j=2;j<=d;j++){ if(arr[i]%j==0) { flag=1; break; } } if(flag==0) { tem=arr[i]; while(tem) { r=tem%10; sum=sum+r; tem=tem/10; } d=int(sum/2); for(j=2;j<=d;j++) { if(sum%j==0){ flag1=1; break; } } if(flag1==0){ cout<<arr[i]<<" "; } } } }
输出结果
如果我们运行上面的程序,那么它将生成以下输出
2 3 11 25 12 7