PAT 甲级 1060 Are They Equal
1060. Are They Equal (25)
时间限制
50 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*105 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.
Input Specification:
Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.
Output Specification:
For each test case, print in a line "YES" if the two numbers are treated equal, and then the number in the standard form "0.d1...dN*10^k" (d1>0 unless the number is 0); or "NO" if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.
Note: Simple chopping is assumed without rounding.
Sample Input 1:
3 12300 12358.9Sample Output 1:
YES 0.123*10^5Sample Input 2:
3 120 128Sample Output 2:
NO 0.120*10^3 0.128*10^3这道题目写的真窝火，所以刚AC，就跑来写博客，好鄙视一下题目。
题目的给定数字会有这样的情况
00234.34000
要把他转换成234.34
另外0.0001科学技术法应该是0.1*10^-3，而不是0.0001*10^0
当n大于数字的个数的时候，要补零
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <string>
using namespace std;
pair<string,int> m;
string a,b;
string aa,bb;
int n;
pair<string,int> fun(string a)
{
int len=a.length();
int begin,end;
int tag1=0,tag2=0;
for(int i=0;i<len;i++)//去除前缀的0和后缀的0
{
if(!tag1&&(a[i]!='0'||(a[i]=='0'&&a[i+1]=='.')))
begin=i,tag1=1;
if(tag2==1&&a[i]!='0')
end=i;
if(tag2==0)
end=i;
if(a[i]=='.')
tag2=1;
}
string c=a.substr(begin,end-begin+1);
int i;
for(i=0;c[i];i++)//找到小数点的位置
if(c[i]=='.')
break;
int j;
for(j=0;c[j];j++)//找到第一个不是0的数字位置，即科学技术法中小数点应该放在哪个位置之前，
{
if(c[j]!='0'&&c[j]!='.')
break;
}
string d="0.";int num=0;
for(int ii=j;c[ii];ii++)
{
if(c[ii]=='.') continue;
d+=c[ii];
num++;
if(num>=n)
break;
}
for(int i=0;i<n-num;i++)//补0
d+='0';
int k=i-j;//小数点位置的变化，就是指数
if(k<0) k++;
m.first=d;
m.second=k;
return m;
}
int main()
{
cin>>n>>a>>b;
pair<string,int> mm1,mm2;
mm1=fun(a);
mm2=fun(b);
if(mm1==mm2)
cout<<"YES "<<mm1.first<<"*10^"<<mm1.second<<endl;
else
cout<<"NO "<<mm1.first<<"*10^"<<mm1.second<<" "<<mm2.first<<"*10^"<<mm2.second<<endl;
return 0;
}