PAT A 1060. Are They Equal (25)
题目
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 12358.9
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
求两个数在n位有效位时是否相等。
注意：
1、转换后小数点后第1位应当为非0数，这个数字可出现在原数的小数点之后，如0.00123；
2、原有数据可能长度不足，如0.00123取5位有效数字，后面的2位应认为是0；
3、注意计算指数时，系数为正和系数为负计算式有区别
代码：
#include <iostream>
#include <string>
using namespace std;
int Transform(string &s,int n); //转换格式，s为输入数字的字符串结构，转换后取有效数字部分，0的话补足够位数的0；返回指数
void Put_out(string s,int exp,int n); //输出数据
int main()
{
int n;
string s1,s2;
cin>>n>>s1>>s2; //输入
int exp1,exp2; //转换
exp1=Transform(s1,n);
exp2=Transform(s2,n);
if(exp1!=exp2) //指数不等，直接跳出
{
cout<<"NO ";
Put_out(s1,exp1,n);
cout<<" ";
Put_out(s2,exp2,n);
return 0;
}
else //指数相等
{
for(int i=0;i<n;i++) //判断有效位范围内是否相等
{
if(s1[i]!=s2[i]) //存在不相等的有效位
{
cout<<"NO ";
Put_out(s1,exp1,n);
cout<<" ";
Put_out(s2,exp2,n);
return 0;
}
} //有效位都相等
cout<<"YES ";
Put_out(s1,exp1,n);
return 0;
}
return 0;
}
void Put_out(string s,int exp,int n) //格式化输出
{
int i;
cout<<"0.";
for(i=0;i<n;i++)
cout<<s[i];
cout<<"*10^"<<exp;
}
int Transform(string &s,int n) //转换格式
{
int i,j;
int pos_point=s.size(); //小数点位置
int pos_begin=s.size(); //第一个非'0'，非'.'的有效字符的位置，
for(i=0;i<s.size();i++)
if(s[i]=='.')
{
pos_point=i;
break;
}
for(i=0;i<s.size();i++)
if(s[i]!='0'&&s[i]!='.')
{
pos_begin=i;
break;
}
if(pos_begin==s.size()) //没有有效位，即数字为0
{
s="0";
for(i=0;i<n;i++)
s+="0";
return 0;
}
for(j=0,i=pos_begin;j<n+2;i++) //转化有效数据
{
if(i<s.size()) //数据在原有效数据段内
{
if(s[i]!='.')
s[j++]=s[i];
}
else if(j<s.size()) //超出范围，但位数不到原数据长度
s[j++]='0';
else //超出原数据长度，n大于输入数据的位数
{
s+="0";
j++;
}
}
if(pos_point-pos_begin>=0) //指数大于0
return pos_point-pos_begin;
else //指数小于0，！！！注意小于零时指数计算是不同的
return pos_point-pos_begin+1;
}