A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1691    Accepted Submission(s): 494 Problem Description Given two positive integers a and b,find suitable X and Y to meet the conditions: X+Y=a Least Common Multiple (X, Y) =b   Input Input includes multiple sets of test data.Each test data occupies one line,including two positive integers a(1≤a≤2*10^4),b(1≤b≤10^9),and their meanings are shown in the description.Contains most of the 12W test cases.   Output For each set of input data,output a line of two integers,representing X, Y.If you cannot find such X and Y,output one line of "No Solution"(without quotation).   Sample Input 6 8 798 10780   Sample Output No Solution 308 490   Source 2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）   Recommend wange2014  

大体题意：

给你a和b，让你求出X和Y，使得X + Y = a    lcm(x,y) = b

思路：

看数据范围肯定不能进行暴力枚举了！

令gcd(x,y) = g;

那么

g * k1 = x;

g * k2 = y;

因为g 是最大公约数，那么k1与k2 必互质！

=>   g*k1*k2 = b

=>   g*k1 + g * k2 = a;

所以k1 * k2 = b / g;

k1 + k2 = a/g;

因为k1与k2 互质！

所以k1 * k2 和 k1 + k2 也一定互质（一个新学的知识点= = ）

所以a/g 与b/g也互质！

那么g 就是gcd(a,b);

所以我们得出一个结论：   gcd(x,y) == gcd(a,b);;

所以x + y 与 x * y都是已知的了，解一元二次方程即可！

#include <bits/stdc++.h> using namespace std; typedef long long ll; ll gcd(ll a,ll b){ return !b ? a : gcd(b, a%b); } int main(){ ll a,b; while (~scanf("%lld %lld",&a, &b)){ int g = gcd(a,b); ll B = b * g; ll q = a*a - 4LL * B; if (q < 0) { puts("No Solution"); continue; } bool ok = 1; ll tmp = sqrt(q); if (tmp * tmp != q) ok = 0; ll a1 = (a + tmp); ll a2 = (a - tmp); if (a1 & 1 || a2 & 1) ok = 0; ll ans1 = min(a1,a2); ll ans2 = max(a1,a2); if (ok)printf("%lld %lld\n",ans1/2,ans2/2); else puts("No Solution"); } return 0; }