1617 Finding Bovine Roots 解题报告

 

Finding Bovine Roots

时间限制(普通/Java):1000MS/10000MS     运行内存限制:65536KByte
总提交: 4            测试通过: 3

http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=1617

描述

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.

 

输入

Line 1: A single integer, L

Line 2: L digits (with no spaces)

 

输出

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

样例输入

 

3
123

 

样例输出

 

17

 

提示

Be careful of floating point arithmetic. It engenders rounding errors that can be surprising.

Explanation of the sample:

sqrt(17) ~= 4.1231056256176605498214098559740770251472

 

说来惭愧 这个题目是看着别人代码理解的，从topsky那里的代码理解的，感谢他首先，一位热心且实力派的ACM爱好者。

废话不多了，看看代码你们就懂了也就。

#include<stdio.h> #include<math.h> int main() { int n,m,t,i,tt,z; double x,c; while(scanf("%d",&n)!=EOF) { t=1; scanf("%d",&m); for(i=1;i<=n;i++) t*=10; x=(double)m/t; while(1) { c=(int)(x*x+0.555); c=sqrt(c); z=(int)c; tt=(c-z)*t; if(tt==m) { printf("%d/n",(int)(x*x+0.555));//这里模拟，那么原先的数SQRT肯定 x*x肯定接近原来的数 故这里+0.555 其实加小点也够自己掌握 break; } x+=1; } } return 0; }