裸题

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 20   Accepted Submission(s) : 5

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Problem Description

Here is no naked girl nor naked runners, but a naked problem: you are to find the K-th smallest element in the set of all irreducible fractions p/q,with 0<p<q<=N.

Input

The first line of input is an integer T(T<=1000), indicating there are T cases.
For each cases, there is one line with two positive integers N and K(1<=K<N<=100000).

Output

For each case, output one line containing the answer.

Sample Input

45 15 25 35 4

Sample Output

1/51/41/32/5

//题意是求分子分母都小于N的第k小的分数。由1/n<1/(n-1) ,2/n>1/(n-1) ,k1/n1>(k1+k2)/(n1+n2)  (n1>n2,k1>k2) 得出

所以通过这个构建子树，从 1/2开始先向左扩张，找到最小的(即 1/n)，然后回退找到父亲那里就是次小的，然后搜右子树，。。

 下面用栈来模拟查找这个过程。

#include<cstdio>const int size=100000;  //栈的最大深度int pa[size],qa[size],pb[size],qb[size]; //栈中当前两个分数的分子分母int s[size]; //栈的当前状态int main(){    int t,n,k,p,q;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        pa[0]=s[0]=0;  //初始化分数和状态        qa[0]=pb[0]=qb[0]=1;        int i=0;  //栈的深度        while(1)  //循环求解        {            p=pa[i]+pb[i];            q=qa[i]+qb[i];  //依栈顶的两个分数计算出来的子节点的分数            if(q>n) i--;  //分母超出n的范围 退栈            else if(s[i]==0) //把当前子节点分数当作右分数            {                pa[i+1]=pa[i];//当前左分数压栈                qa[i+1]=qa[i];                pb[i+1]=p;    //当前右分数压栈                qb[i+1]=q;   //进入下一步                s[i]++;  //新的栈顶元素初始化状态                s[i+1]=0; //栈的深度加1                i++;            }            else            {   //把当前子节点分数当作左分数                k--;  //找到一个最小分数                if(k==0) break;                pa[i]=p;                qa[i]=q;                s[i]=0;            }        }        printf("%d/%d\n",p,q);    }    return 0;}