Problem 49 Prime permutations （set + vector）

Prime permutations

Problem 49

The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.

There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.

What 12-digit number do you form by concatenating the three terms in this sequence?

Answer: 296962999629 Completed on Mon, 31 Oct 2016, 10:42

代码：

#include<bits/stdc++.h>
using namespace std;
int is_prime(int n)
{
if (n <= 1)
{
return 0;
}
else if (n == 2 || n == 3)
{
return 1;
}
else if (n % 2 == 0)
{
return 0;
}
else
{
for (int i=3;i<=(int)sqrt((float)n);i++)
{
if (n % i == 0)
{
return 0;
}
}
return 1;
}
}

int same_digits(int a,int b)
{
vector<int> va(4);
vector<int> vb(4);

while(a != 0 && b != 0)
{
va.push_back(a%10);
vb.push_back(b%10);
a /= 10;
b /= 10;
}

sort(va.begin(),va.end());
sort(vb.begin(),vb.end());

return va == vb;
}

int main()
{
vector<int> primes; //保存所有四位数的质数
for (int n=1000;n<=9999;n++)
{
if(is_prime(n)) primes.push_back(n);
}

int num = primes.size(); //质数个数
for (int i=0;i<num-2;i++) //遍历所有质数
{
for (int j=i+1;j<num;j++)
{
if(same_digits(primes[i],primes[j])) //判断两个质数是否含有相同的数字
{
int diff = primes[j] - primes[i];
int last = primes[j] + diff; //等差数列第三个数
vector<int>::iterator it;
it = find(primes.begin(),primes.end(),last); //查找第三个数 last 是否在primes中
if (it != primes.end()) //在primes中
{
if (same_digits(primes[i],last)) //含有相同的数字
{
cout<<primes[i]<<" "<<primes[j]<<" "<<last<<endl;
}
}

}
}
}
return 0;
}