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...d_N*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 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample 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;
}