Some projects--Xiao Ming's punishment

The topic description Xiao Ming and Xiao Hui Quarrel, Xiao Ming wants to see small Hui and beg to be forgiven. Xiao Hui gave Xiao Ming two number M and N (1<m<1000 and m<n<2000), between M to n Prime (the range contains M and N), the maximum number of adjacent two prime difference is small Hui decided not to see Xiao Ming's days, xiaoming very want to know a few days later to see small Hui, Clever you help xiaoming?

Input

Input m and N

Output

Maximum number of adjacent two prime difference

Sample input

10 30

Sample output

6

Tips

The prime number between 10 and 30 is 11 13 17 19 23 29, and the maximum of the adjacent two prime difference is 29-23 = 6

Answer

#include <iostream>using namespace Std;bool is (int n) {    int m=0,i;    for (I=1; i<n; i++)        if (n%i==0)            m++;    if (m==1)        return true;    else        return false;} int main () {    int a,b,m[1000],i,n=0,max=0;    cin>>a>>b;    For (I=a, i<=b; i++)        if (is (i))        {            m[n++]=i;        }    for (i=0; i<n-1; i++)    {        if (M[i+1]-m[i]>max)            max=m[i+1]-m[i];    }    Cout<<max;;}

Some projects--Xiao Ming's punishment