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：
1 #include<bits/stdc++.h> 2 //#include<regex> 3 #define db double 4 #include<vector> 5 #define ll long long 6 #define vec vector<ll> 7 #define Mt vector<vec> 8 #define ci(x) scanf("%d",&x) 9 #define cd(x) scanf("%lf",&x) 10 #define cl(x) scanf("%lld",&x) 11 #define pi(x) printf("%d ",x) 12 #define pd(x) printf("%f ",x) 13 #define pl(x) printf("%lld ",x) 14 #define MP make_pair 15 #define PB push_back 16 #define inf 0x3f3f3f3f3f3f3f3f 17 #define fr(i,a,b) for(int i=a;i<=b;i++) 18 const int N=1e3+5; 19 const int mod=1e9+7; 20 const int MOD=mod-1; 21 const db eps=1e-18; 22 const db PI=acos(-1.0); 23 using namespace std; 24 map<ll,int> mp; 25 int main(){ 26 //freopen("data.in","r",stdin); 27 //freopen("data.out","w",stdout); 28 ll a,b; 29 for(int i=0;i<=20005;i++){ 30 ll x=1ll*i*i; 31 mp[x]=i; 32 } 33 while(scanf("%lld%lld",&a,&b)==2){ 34 bool ok=0; 35 ll i=__gcd(a,b); 36 // for(ll i=1;i*i<=a;i++) { 37 if(a%i==0&&b%i==0){ 38 ll bb=b/i,aa=a/i,y=aa*aa-4*bb; 39 if(y<0||!mp.count(y)||(aa+mp[y])%2==1){printf("No Solution ");continue;} 40 ll ans1=(aa+mp[y])/2; 41 ll ans2=(aa-mp[y])/2; 42 ans1*=i,ans2*=i; 43 ll cnt1=a-ans1,cnt2=a-ans2; 44 if(cnt1*ans1/__gcd(cnt1,ans1)==b){ 45 printf("%lld %lld ",cnt1,ans1); 46 } 47 else if(cnt2*ans2/__gcd(cnt2,ans2)==b){ 48 printf("%lld %lld ",cnt2,ans2); 49 } 50 } 51 // } 52 53 } 54 return 0; 55 }