POJ 3175 Finding Bovine Roots

最新推荐文章于 2019-07-22 22:11:00 发布

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最新推荐文章于 2019-07-22 22:11:00 发布

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分类专栏： 2017.4 第四周--学习图论文章标签： poj

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本文探讨了一种特殊的平方根计算方法，即“Bovine平方根”。目标是在给定长度L的数字串的情况下，找到最小的整数，使得该整数的Bovine平方根的小数部分开始匹配给定的数字串。

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Finding Bovine Roots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4546		Accepted: 931

Description

The cows are trying to find their roots. They are not so smart as humans when they find square roots, the cows lose the integer portion of their square root calculation. Thus, instead of calculating the square root of 2 to be 1.4142135623730950488016887242096980785696, they deduce that it is 0.4142135623730950488016887242096980785696. The square root of 16 is calculated to be 0 (obviously an error).

The erroneous ciphering does, however, lead to an interesting question. Given a string of digits of length L (1 <= L <= 9), what is the smallest integer whose bovine square root decimal part begins with those digits?

By way of example, consider the string "123". The square root of 17 is approximately 4.1231056256176605498214098559740770251472 so the bovine square root is: 0.1231056256176605498214098559740770251472 whose decimal part (just after the period) starts with 123. It turns out that 17 is the smallest integer with this property.

Input

Line 1: A single integer, L

Line 2: L digits (with no spaces)

Output

Line 1: A single integer that is the smallest integer whose bovine square root starts with the supplied string

Sample Input

3
123

Sample Output

这道题目是给定一个指定数位L的整数N。寻找一个最小的整数，对他求平方根，令小数点后的L位与N相等。

第一次做的时候暴力列举的时候TL，显然方法不对。采用枚举的方法，直接将输入的整数化作小数，每一次加1，递增。看它的平方取整之后与它+(1e+L)平方取整的大小，以此作为+1的结束条件。类似于sqrt(16) 与sqrt(17) 两者的平方根取整都为4，但是他们本身的值不同。

#include<cstdio>
#include<math.h>
#include <iostream>
#define eps 1e-8
using namespace std;
typedef long long ll;
double a[11]={1,1e-1,1e-2,1e-3,1e-4,1e-5,1e-6,1e-7,1e-8,1e-9,1e-10};

int main(){
    int n;
    int m;
    double p,q,x,y;
    while(scanf("%d",&n)!=EOF){
        scanf("%d",&m);
        p=m*a[n];  q=(m+1)*a[n];
        x=y=0;
        while(x==y){
            p+=1; q+=1;
            x=(ll)(p*p);  y=(ll)(q*q-eps);//使用long long 取整，防止溢出
        }
        printf("%.f\n",(y));
    }
    return 0;
}