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\n", x1*g/2, x2*g/2); } return 0; }