hdu 5974 A Simple Math Problem
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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 8798 10780
Sample Output
No Solution308 490思路:
由题意可知:
lcm(x,y)=b;
因为 x*y =gcd(x,y)*lcm(x,y);
所以 x*y =gcd(x,y)*b;
因为 gcd(a,b)=gcd(x,y);
所以 x*y=b*gcd(a,b);
联立 x+y=a;
解个方程就行 ;#include <cstdio>#include <cstring>#include <cmath>int gcd (int m,int n){int flag;while(n>0){flag=m%n;m=n;n=flag;}return m;}int main (){int a,b;while(scanf("%d%d",&a,&b)!=EOF){int i,j;int c=gcd(a,b);int x,y;int delta= a*a-4*b*c;if(delta<0){printf("No Solution\n");continue;}int flag= sqrt(delta);if(flag*flag!=delta||(a-flag)%2!=0){printf("No Solution\n");continue;}x=(1*a-sqrt(delta))/2;y=a-x;printf("%d %d\n",x<y?x:y,x<y?y:x);}return 0;}
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