srm 535 div2 250&500

250：題意：知道A+B，B-C，A-B ,B+C ，的值求ABC的值，很水吧- -

做法是,先什麼都不考慮就把ABC算出來，然後回過頭驗證是否符合

：

AminusB = A - B BminusC = B - C AplusB = A + B BplusC = B + C

或者是枚舉，ABC，3重循環。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
 
using namespace std;
 
class FoxAndIntegers {
	public: 
	vector <int>get(int AminusB, int BminusC, int AplusB, int BplusC) 
	{
 		 int A, B, C;
 		 B = (BminusC + BplusC)/2;
 		 A = (AminusB + AplusB)/2;
 		 C = BplusC - B;
 		 bool pos = true;
 		 vector<int> ans;
 		 if ( A-B != AminusB||A+B != AplusB||B-C != BminusC||C+B != BplusC ) pos = false;
 		 if ( pos ) 
 		 {
	    	 ans.push_back(A);
 	    	 ans.push_back(B);
	   	     ans.push_back(C);
	      }
	  	return ans;
    }
};
           

500pt：

題意：

知道兩數的最大公約數和最小公倍數，球這兩個數。關鍵是此題資料比較大10^12.

根據

L=(A*B)/G

兩邊同時再除以一個G

L/G = (A/G) * (B/G)

這兩個數也是互質的

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
 
using namespace std;
 
class FoxAndGCDLCM {
public:
  long long gcd(long long a, long long b) 
  {
    if ( a == 0)  return b;
    return gcd(b%a, a);
  }
  long long FoxAndGCDLCM::get(long long G, long long L)
  {
    	if ( L%G != 0 ) return -1;
 	    long long C = L/G, ans = 1LL<<60;
 	    for ( long long i =1; i*i <= C; i++ ) 
 	    {
   			 if ( C%i == 0 ) 
   			 {
     	    	 long long j = C/i;
    	    	 if ( gcd(i,j) == 1 )
    	    	 {
       				 ans = min(ans, G*(i+j));
     			 }
    		}
  		}
 	   if ( ans == 1LL<<60 ) ans =-1;
 	   return ans;
	}
};