UVA 10140

meaning of the problem:

Given two int L and U, find 2 pair of prime number. The first pair should be adjacent prime number the difference between whom is the smallest and that of the other pair should be the largest.

Outline：

Since the range for U and L is 1~(2^32-1), difference between U and L will not exceed 1，000,000， we can easily get prime number in 2^16 using sieve method. And then use the prime numbers we got to kick out non prime number in U and L,  

Tips: the function cal is important so that we will not get TLE.

if you get run time error, you can check whether you used s- u in the flag_in_range.

#include<stdlib.h>  #include<cstring>  #include<cmath>  #include<stack>  #include<queue>  #include<algorithm>  #include <cstdio>  #define MAX(a,b) ((a)>(b)?(a):(b))   #define maxfacprime 65536 // int^1/2  using namespace std;  int prime[50000];  bool flag[70000];  int count_p = 0;    void findprime()  {      int i, j;      memset(flag, true, sizeof(flag));      for (i = 2; i <= maxfacprime; i++)      {          if (flag[i] == true) prime[++count_p] = i;          for (j = 1; j <= count_p&&i*prime[j] <= maxfacprime; j++)          {              flag[i*prime[j]] = false;              if (i%prime[j] == 0)                  break;          }      }  }  long long int prime_in_range[1000000];  bool flag_in_range[1000000];  long long int cal(long long int n,long long int lim)  {      long long int k = lim / n;      if (k*n == lim)          return lim;      else return n*(k + 1);  }  int main()  {      //freopen("in.txt", "r", stdin);      //freopen("out.txt", "w", stdout);      //cout << maxfacprime;      long long int u, v;      findprime();      while (cin >> u >> v)      {          if (u == v)          {              printf("There are no adjacent primes.\n");              continue;          }              int count_prime_in_range = 0;          memset(flag_in_range, false, sizeof(flag_in_range));          long long int m = (int)sqrt(v) + 1;          for (int i = 1; i <= count_p; i++)          {              if (prime[i] > m)                  break;  //          long long int j = 2;          /*  while (j*prime[i] < u)                 j++;*/              for (long long int s = cal(prime[i],u); s <= v; s += prime[i])              {                  if (s == prime[i])                      continue;                  flag_in_range[s-u] = true;              }          }          for (long long int k = u; k <= v; k++)          {              if (k == 1)                  continue;              if (flag_in_range[k-u])                  continue;              prime_in_range[count_prime_in_range++] = k;          }          long long int small_dis = 1000000;          long long int large_dis = 0;          long long int s1, s2;          long long int l1, l2;          for (int i = 0; i < count_prime_in_range - 1; i++)          {              long long int dis = prime_in_range[i + 1] - prime_in_range[i];              if (small_dis > dis)              {                  small_dis = dis;                  s1 = prime_in_range[i];                  s2 = prime_in_range[i+1];              }              if (large_dis < dis)              {                    large_dis = dis;                  l1 = prime_in_range[i];                  l2 = prime_in_range[i + 1];              }          }          if (count_prime_in_range <= 1)              printf("There are no adjacent primes.\n");          else printf("%lld,%lld are closest, %lld,%lld are most distant.\n", s1, s2, l1, l2);        }      return 0;  }