Sgu 499 greatest common divisor"

Question address: http://acm.sgu.ru/problem.php? Contest = 0 & problem = 499

Analysis logic: Directly calculating the maximum common number of two pairs will time out. Let's look at the problem from another perspective. We can find all the approximate number of input numbers. If a number appears twice or more, then, the most satisfying condition is ggcd.

Then it is similar to the sieve method to find the approximate number, so you can find 1 ~ In N, each number cannot exceed all the approx. Of its own square root. Then the one that is bigger than the square root can be used for Division (the exact number of vertices is special ).

Then, instead of brute force, we can directly find the one that appears twice from the largest one. We can set a maxn to record it. Then we can compare the number of the second appearance with maxn.

First look at the Code:

#include<iostream>#include<vector>#include<cmath>//#include<cstdio>using namespace std;#define  N 1000005vector<int>  v[N+1];//int cnt[N+1]={0};bool b[N+1]={0};void pre(){    int len=sqrt(N);   for(int i=1;i<=len;i++)     for(int j=i*i;j<=N;j+=i)        v[j].push_back(i);}int maxn=1;void check(int n){   if(b[n])   {       maxn=n>maxn?n:maxn;   }   b[n]=1;}int main(){   pre();   int size;   cin>>size;   for(int i=0;i<size;i++)      {         int n;         cin>>n;               if(n<=maxn)  continue;         for(int j=0;j<v[n].size();j++)         {              check(v[n][j]);             if(v[n][j]*v[n][j]!=n)              check(n/v[n][j]);         }//            int sqrtn=sqrt(n);//            for(int j=1;j<=20&&j<=sqrtn;j++)//          {//              if(n%j==0)//              { check(j);////                if(j*j!=n) check(n/j);////              }//          }      }//    for(int i=1000000;i>0;i--)//       if(cnt[i]>=2)//       {//           cout<<i<<endl;//           break;//       }     cout<<maxn<<endl; }

This will time out, because each number has a factor of 1, and the general number has a factor of 2. When we create a table, we choose not to input the first 20 orders, then, check the first 20 items for the input number (after all, only 10 million items are smaller than the data range of 100 million.

Then, SQRT (n) should not appear in for (INT I = 0; I, <SQRT (n); I ++, otherwise, the computation is performed every time you make a decision. Because this TLE is used several times.

When TLE is always stuck in test 20, change all CIN cout to scanf printf. You can run test 38 to improve the fur.

The Code is as follows:

#include<iostream>#include<vector>#include<cmath>//#include<cstdio>using namespace std;#define  N 1000005vector<int>  v[N+1];//int cnt[N+1]={0};bool b[N+1]={0};void pre(){    int len=sqrt(N);   for(int i=21;i<=len;i++)     for(int j=i*i;j<=N;j+=i)        v[j].push_back(i);}int maxn=1;void check(int n){   if(b[n])   {       maxn=n>maxn?n:maxn;   }   b[n]=1;}int main(){   pre();   int size;   cin>>size;//   scanf("%d",&size);   for(int i=0;i<size;i++)      {         int n;         cin>>n;       // scanf("%d",&n);          if(n<=maxn)  continue;         for(int j=0;j<v[n].size();j++)         {              check(v[n][j]);             if(v[n][j]*v[n][j]!=n)              check(n/v[n][j]);         }            int sqrtn=sqrt(n);            for(int j=1;j<=20&&j<=sqrtn;j++)          {              if(n%j==0)              { check(j);                if(j*j!=n) check(n/j);              }          }      }//    for(int i=1000000;i>0;i--)//       if(cnt[i]>=2)//       {//           cout<<i<<endl;//           break;//       }     cout<<maxn<<endl;  // printf("%d\n",maxn);}