hdu5974 or SDKD 2017 Spring Team Training C 第D题( 韦达定理 解方程)

A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1634    Accepted Submission(s): 460

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

 

mdzz 

这个题 煞费脑筋- -     到结束没有退出gcd(x,y)==gcd(a,b); ╮(╯▽╰)╭   还是数学不好

12w 数据 扫一遍 肯定超时; 一个for 也超时;  

 解题思路:

x+y= a;

lcm(x,y)=b;  即 x*y=b*gcd(x,y);

联立方程 有 x²-a*x+b*gcd(x,y)==0  解这个方程的 解就可以

现在推gcd(x,y)=gcd(a,b);   

另gcd(x,y)= k; 

 得出 gcd(x,y)=gcd(a,b);

接着就是解 二元一次方程的问题

detal=a²-4*b*gcd(x,y);

detal<0 无解

大于零则  x=(a-sqrt(detal))/2  y=a-x;  注意 x要小于y  有两个解 只输出第一个  x<y的   并且要注意

sqrt(detal) 为整数才可以 

#include<stdio.h>#include<math.h>/*gcd(x,y)=gcd(a,b)*/int gcd(int x,int y){    if(y==0) return x;    else return gcd(y,x%y);}int main() {    int a,b;    int x,y,i,j;    int low,high;    while(~scanf("%d %d",&a,&b))    {        int k=gcd(a,b);        int detal=a*a-4*b*k;        if(detal<0)        {            printf("No Solution\n");            continue;        }        else        {            int  s=sqrt(detal);            if(s*s!=detal||(a-s)%2!=0)            {                printf("No Solution\n");                    continue;            }            x=(a-s)/2;            y=a-x;            x<y?x:y;            y=a-x;            printf("%d %d\n",x,y);        }    }    return 0;}