【HDOJ5974】A Simple Math Problem（构造，解方程）

题意：给定A与B，要求构造出一组X,Y，使得X+Y=A，lcm(X,Y)=B

A<=2e4，B<=1e9

思路：A的范围较小，考虑以A为突破口

枚举A的约数k，复杂度O(sqrt(A))

设X=pk，y=qk，p与q互质

原方程转化：

(p+q)k=a     ——>p+q=a/k

pqk=b 　　——>pq=b/k

p,q即为方程x^2-a/k*x+b/k=0的一组正整数解

解得p,q后X=pk,y=qk

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N   2100000
 #define MOD 1000000007
 #define eps 1e-8
 #define pi acos(-1)

 ll A,B,s1,s2;

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 ll gcd(ll x,ll y)
 {
     if(y==) return x;
     return gcd(y,x%y);
 }

 int isok(ll k)
 {
     if(A%k) return ;
     if(B%k) return ;
     ll a=;
     ll b=-A/k;
     ll c=B/k;
     ll x=b*b-a*c*;
     if(x<) return ;
     ll delta=(long long int)(sqrt(x)+eps);
     //printf("%lld %lld %lld %lld %lld %lld\n",a,b,c,x,delta,k);
     if(delta*delta!=x) return ;
     ll t=-b+delta;
     if(t%(a*)) return ;
     ll x1=t/(a*)*k;
     t=-b-delta;
     if(t%(a*)) return ;
     ll x2=t/(a*)*k;
     //printf("%lld %lld\n",x1,x2);
     if(x1&&x2&&gcd(x1,x2)==k)
     {
         if(x1>x2) swap(x1,x2);
         s1=x1;
         s2=x2;
         return ;
     }
     return ;
 }

 int main()
 {
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    while(scanf("%I64d%I64d",&A,&B)!=EOF)
    {
             s1=s2=;
            for(ll i=;i<=sqrt(A+eps);i++)
             if(A%i==)
          {
              if(isok(i)) break;
              if(isok(A/i)) break;
          }
         //printf("%I64d %I64d\n",s1,s2);
         if(s1+s2==) printf("No Solution\n");
          else
          {
              if(s1>s2) swap(s1,s2);
              printf("%I64d %I64d\n",s1,s2);
          }

    }

 }