hdoj 5974 A Simple Math Problem

题目链接：A Simple Math Problem

题目大意：给你一个a和b，现在要求x+y = a，lcm(x,y) = b,输出x,y

题目思路：数据范围太大，暴力不可解，我们假设s=gcd(x,y) 则 i∗s+j∗s=a,(i∗s∗j∗s)/s=b即(i+j)∗s=a,(i∗j)∗s=b,易知i,j互质，故i+j与i*j互质，所以gcd(a,b)==s，这样我们就求出了s,然后i，j两个未知数，两个方程，可解，考虑一下无解和delta的情况，然后就可出了，具体看代码

#include <map>#include <set>#include <cmath>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;int main(){    ll a,b;    while(~scanf("%lld%lld",&a,&b)){        ll x = a/__gcd(a,b);        ll y = b/__gcd(a,b);        if(x*x-4*y < 0) puts("No Solution");        else{            ll sq = sqrt(x*x-4*y);            if(sq*sq != x*x-4*y) puts("No Solution");            else{                ll ans1 = x+sq;                ll ans2 = x-sq;                if(ans1 > ans2) swap(ans1,ans2);                if(ans1%2 == 1) puts("No Solution");                else printf("%lld %lld\n",ans1/2*__gcd(a,b),ans2/2*__gcd(a,b));            }        }    }    return 0;}