HDU 5974 数学

A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2891    Accepted Submission(s): 907

Problem DescriptionGiven 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

 

题意：给你 x+y=a, lcm(x,y)=b 求 a，b

思路：

易得：x/gcd+y/gcd=a/gcd, x/gcd*y/gcd=b/gcd; 令i=x/gcd, j=y/gcd; 即i*gcd+j*gcd=a,i*j*gcd=b ==> gcd(a,b)=gcd(x,y);解个方程即可。

PS：刚开始用的枚举gcd的方法，复杂度1e7，但是一直WA？？？

代码：

 #include<bits/stdc++.h>
 //#include<regex>
 #define db double
 #include<vector>
 #define ll long long
 #define vec vector<ll>
 #define Mt  vector<vec>
 #define ci(x) scanf("%d",&x)
 #define cd(x) scanf("%lf",&x)
 #define cl(x) scanf("%lld",&x)
 #define pi(x) printf("%d\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 #define MP make_pair
 #define PB push_back
 #define inf 0x3f3f3f3f3f3f3f3f
 #define fr(i,a,b) for(int i=a;i<=b;i++)
 const int N=1e3+;
 const int mod=1e9+;
 const int MOD=mod-;
 const db  eps=1e-;
 const db  PI=acos(-1.0);
 using namespace std;
 map<ll,int> mp;
 int main(){
 //freopen("data.in","r",stdin);
 //freopen("data.out","w",stdout);
     ll a,b;
     for(int i=;i<=;i++){
         ll x=1ll*i*i;
         mp[x]=i;
     }
     while(scanf("%lld%lld",&a,&b)==){
         bool ok=;
         ll i=__gcd(a,b);
 //        for(ll i=1;i*i<=a;i++) {
             if(a%i==&&b%i==){
                 ll bb=b/i,aa=a/i,y=aa*aa-*bb;
                 if(y<||!mp.count(y)||(aa+mp[y])%==){printf("No Solution\n");continue;}
                 ll ans1=(aa+mp[y])/;
                 ll ans2=(aa-mp[y])/;
                 ans1*=i,ans2*=i;
                 ll cnt1=a-ans1,cnt2=a-ans2;
                 if(cnt1*ans1/__gcd(cnt1,ans1)==b){
                     printf("%lld %lld\n",cnt1,ans1);
                 }
                 else if(cnt2*ans2/__gcd(cnt2,ans2)==b){
                     printf("%lld %lld\n",cnt2,ans2);
                 }
             }
 //        }

     }
     return ;
 }