A simple problem 分类： 哈希 HDU 2015-08-06 08:06 1人阅读 评论(0) 收藏

A simple problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3702 Accepted Submission(s): 1383

Problem Description

Zty很痴迷数学问题.。一天，yifenfei出了个数学题想难倒他，让他回答1 / n。但Zty却回答不了^_^. 请大家编程帮助他.

Input

第一行整数T，表示测试组数。后面T行，每行一个整数 n (1<=|n|<=10^5).

Output

输出1/n. (是循环小数的,只输出第一个循环节).

Sample Input

4

2

3

7

168

Sample Output

0.5

0.3

0.142857

0.005952380

Author

yifenfei

Source

HDU 2008-10 Programming Contest

一道有意思的题,哈希思想,不过在初始化的时候有点小技巧可以防止超时.

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker, "/STACK:102400000")
#define WW freopen("output.txt","w",stdout)

const int MAX = 1e5+10;

bool vis[MAX*10];

int main()
{
    int T;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);

        if(n<0)
        {
            printf("-");
            n=-n;
        }
        if(n==1)
        {
            printf("1");
        }
        else
        {
            printf("0.");
            memset(vis,false,(n*10)*sizeof(vis[0]));//不必全部初始化
            m=10;
            vis[10]=true;
            while(1)
            {
                printf("%d",m/n);
                vis[m]=true;
                m=m%n;
                if(m==0)
                {
                    break;
                }
                m*=10;
                if(vis[m])
                {
                    break;
                }
            }
        }
        printf("\n");
    }
    return 0;
}

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