HDU 5974 A Simple Math Problem (解方程)

题意：给定a和b，求一组满足x+y=a && lcm(x, y)=b。

析：x+y = a, lcm(x, y) = b,=>x + y = a, x * y = b * k，其中 k = gcd(x, y)。

然后第一个式子同时除以k，第二个式子同时除以k*k，那么x/k,和y/k是互质的，那么a/k和b/k也是互质的。所以问题就转化成了

x' + y' = a'，x' * y' = b'。然后解方程并判断解的存在即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
  int a, b;
  while(scanf("%d %d", &a, &b) == 2){
    int g = gcd(a, b);
    a /= g;  b /= g;
    int det = a * a - 4 * b;
    if(det < 0){ puts("No Solution");  continue; }
    int t = int(sqrt(det+0.5));
    int x1 = a - t;
    if(t * t != det || x1 < 0 || x1 % 2){ puts("No Solution");  continue; }
    int x2 = a + t;
    printf("%d %d
", x1*g/2, x2*g/2);
  }
  return 0;
}