HDU 5974 A Simple Math Problem （数论，大连区域赛）

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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 <cstdio>#include <cstring>#include <algorithm>#include <cmath>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;}

A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 602 Accepted Submission(s): 233

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 8798 10780

Sample Output

No Solution308 490

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

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