hdu5974数学

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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

x+y=a
lcm(x,y)=b

令
x=nd
y=md

有
gcd(n,m)=1
(n+m)d=a
n*m*d=b
=>
n+m=a/d
n*m=b/d
枚举a的因子d，然后就是韦达定理

#include <bits/stdc++.h>using namespace std;typedef long long LL ;const int N = 20000 ;std::vector<int> fac[N+8] ;void init(){    for(int i = 1 ; i <= N ; i++){        for(int j = i ; j <= N ; j += i) fac[j].push_back(i) ;    }}void gao(LL a , LL b){     std::vector<std::pair<LL , LL> > res ;     for(std::vector<int>::iterator it = fac[a].begin() ; it != fac[a].end() ; it++){        int d = *it ;        if((b%d) != 0) continue ;        LL A = 1;        LL B = -a/d ;        LL C = b/d ;        LL delt = B*B - A*C*4 ;        LL de = (LL)std::sqrt(delt + 0.5) ;        if(de*de != delt) continue ;        if((-B + de) % (A*2) == 0){            LL m = (-B + de) / (A*2) ;            LL n = a/d - m ;            if(std::__gcd(n,m) == 1 && n*m == b/d){                if(n > m) std::swap(n , m) ;                res.push_back(std::make_pair(n*d , m*d)) ;            }        }        if((-B - de) % (A*2) == 0){            LL m = (-B - de) / (A*2) ;            LL n = a/d - m ;            if(std::__gcd(n,m) == 1 && n*m == b/d){                if(n > m) std::swap(n , m) ;                res.push_back(std::make_pair(n*d , m*d)) ;            }        }     }     if(res.empty()) puts("No Solution") ;     else{        std::sort(res.begin() , res.end()) ;        printf("%I64d %I64d\n" , res[0].first , res[0].second) ;     }}int main() {    init() ;    LL a , b ;    while(scanf("%I64d%I64d" , &a , &b) != EOF){       gao(a , b) ;    }    return 0 ;}

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