UVA10976Fractions Again?!（暴力解法）

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继续暴力
先分析，xy关系可以推算出y的最大范围2k
解：x>=y 即1/x<=1/y 由于1/x=1/k - 1/y 代入得1/k<=2/y 即y<=2k
然后枚举y即可求出x，再判断。
It is easy to see that for every fraction in the form 1/k(k>0), we can always nd two positive integersxandy, x >=y, such that:
1/k = 1/x + 1/y
Now our question is: can you write a program that counts how many such pairs of x and y there are for any given k?
Input
Input contains no more than 100 lines, each giving a value of
k(0

#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>#include<iomanip>#include<cstring>#include<vector>#include<iterator>using namespace std;int main(){    int x,y,k,s;    while(scanf("%d", &k)!=EOF)    {        s=0;        for(y=k+1; y<=2*k; y++){            if((k*y)%(y-k)==0){                x=(k*y)/(y-k);                if(x>=y) s++;            }        }        printf("%d\n", s);        for(y=k+1; y<=2*k; y++){            if((k*y)%(y-k)==0){                x=(k*y)/(y-k);                if(x>=y) printf("1/%d = 1/%d + 1/%d\n", k, x, y);            }        }    }    return 0;}

要改注意全改啊。。。别只改一处啊。。。笨蛋。。。

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