hdu 5974 A Simple Math Problem gcd(x,y)=gcd((x+y),lcm(x,y))

题目链接

题意

现有$$x+y=a\lcm(x,y)=b$$找出满足条件的正整数\(x,y\).

\(a\leq 2e5,b\leq 1e9,数据组数12W\).

思路

结论

\(gcd(x,y)=gcd((x+y),lcm(x,y))\)

证明

先证\(gcd(x,y)|gcd((x+y),lcm(x,y))\)

不妨设\(gcd(x,y)=k\)，则有\(k\mid x,k\mid y\)，则有\(k\mid (x+y)\) …①

又\(k\mid x,x\mid lcm(x,y)\)，所以\(k\mid lcm(x,y)\) …②

综合①②有\(k\mid gcd((x+y),lcm(x,y))\).

再证\(gcd((x+y),lcm(x,y))|gcd(x,y)\)

设\(gcd((x+y),lcm(x,y))=k_0\)，则有\(k_0|(x+y)\). 若\(k_0\nmid x\)，则有\(k_0\nmid y\)，否则\(k_0\nmid (x+y)\).

而若\(k_0\nmid x,k_0\nmid y\)，则\(\exists d1,d1\mid k_0,d1\nmid x;\exists d2,d2\mid k_0,d2\nmid y\).而\(\forall d\mid lcm(x,y)，有d\mid x或d\mid y\)，...好像不太对......不会证了

// 待补

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a, b;
LL gcd(LL a, LL b) { return b ? gcd(b, a%b) : a; }
void work() {
    LL k = gcd(a, b);
    LL delta = a*a - 4*k*b;
    if (delta < 0) { printf("No Solution\n"); return; }
    LL ssqrt = sqrt(delta);
    if (ssqrt*ssqrt != delta) { printf("No Solution\n"); return; }
    LL den = a+ssqrt;
    if (den&1) { printf("No Solution\n"); return; }
    LL x = den/2;
    LL y = a-x;
    if (x > y) swap(x, y);
    printf("%lld %lld\n", x, y);
}
int main() {
    while (scanf("%lld%lld", &a, &b) != EOF) work();
    return 0;
}