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  • hdu5974

    A Simple Math Problem

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 1795    Accepted Submission(s): 524


    Problem Description

    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
     
    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
     1 #include <iostream>
     2 #include <cmath>
     3 #include <algorithm>
     4 #include <string>
     5 #include <cstring>
     6 #include <cstdio>
     7 using namespace std;
     8 long long f(long long a,long long b)
     9 {
    10     return (b==0)?a:f(b,a%b);
    11 }
    12 int main()
    13 {
    14     long long a,b,l,ans,t;
    15     while(~scanf("%I64d%I64d",&a,&b))
    16     {
    17         l=f(a,b);
    18         if(a*a-4*b*l<0)
    19             cout<<"No Solution"<<endl;
    20         else
    21         {
    22             ans=(a-sqrt(a*a-4*b*l))/2;
    23             t=a-ans;
    24             if(ans*t/f(t,ans)==b)
    25                 cout<<ans<<' '<<t<<endl;
    26             else
    27               cout<<"No Solution"<<endl;
    28         }
    29     }
    30     return 0;
    31 
    32 }
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  • 原文地址:https://www.cnblogs.com/--lr/p/6854528.html
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