1060 Are They Equal (25 分)
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 0.123×10
5
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 10
100
, 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.d[1]...d[N]*10^k (d[1]>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 0.12010^3 0.12810^3
作者
CHEN, Yue
单位
浙江大学
代码长度限制
16 KB
时间限制
400 ms
内存限制
思路
代码
#include <cstdio>
#include<iostream>
#include <string>
using namespace std;
int n;
string deal(string s,int &e) {
int k = 0;//消除0
while(s.length()>0&&s[0] == '0') {
s.erase(s.begin());
}
if(s[0] == '.') { //0.aabb型
s.erase(s.begin());
while(s.length()>0&&s[0] == '0') {
s.erase(s.begin());//0.000as型
e--;
}
} else {//寻找小数点删除
while(k<s.length()&&s[k]!='.'){
k++;
e++;
}
if(k<s.length())
s.erase(s.begin() + k);//删除小数点
}
if(s.length() == 0)//直接为 0
e = 0;
int num = 0;
k = 0;
string res;//按照精度重组的主体部分
while(num < n){
if(k<s.length()) res+=s[k++];
else res += '0';//精度不够 用0 来凑
num ++;//精度
}
return res;
}
int main() {
string s1,s2,s3,s4;
cin>>n>>s1>>s2;
int e1 = 0;
int e2 = 0;
s3 = deal(s1,e1);
s4 = deal(s2,e2);
if(s3 == s4 && e1 ==e2)
cout<<"YES 0."<<s3<<"*10^"<<e1<<endl;
else
cout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2<<endl;
}