PAT A1060

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 0.12310^5
Sample Input 2:
3 120 128
Sample Output 2:
NO 0.120
10^3 0.128*10^3

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#include "stdio.h"
//#include "math.h"
#include "string.h"
#include "iostream"
//#include "stdlib.h"
#include "algorithm"
using namespace std;
//typedef long long LL;
int n;//有效位数
string deal(string s, int& e){
int k = 0;//s的下标
while (s.length() > 0 && s[0] == '0') {
s.erase(s.begin());//去掉s的前导0
}
if (s[0] == '.') {//去掉前导0后后是小数点,说明s是小于1的小数
s.erase(s.begin());//去掉小数点
while (s.length() > 0 && s[0] == '0') {
s.erase(s.begin());//去掉小数点后非零位前的所有零
e--;
}
}else{//去掉前导零后不是小数点,则找到后面的小数点删除
while (k < s.length() && s[k] != '.') {//寻找小数点
k++;
e++;//只要不碰到小数点就让指数e++
}
if (k < s.length()) {//while结束后k < s.length()说明喷到了小数点
s.erase(s.begin() + k);//把小数点删除
}
}
if (s.length() == 0) {
e = 0;//如果去除前导零后s的长度变为0,说明这个数是0
}
int num = 0;
k = 0;
string res;
while (num < n) {//只要精度还没有达到n
if (k < s.length()) {
res += s[k++];//只要还有数字,就加到res末尾
}else{
res += '0';//否则res末尾添加0
}
num++;
}
return res;
}
int main(){
string s1, s2, s3, s4;
cin >> n >> s1 >> s2;
int e1 = 0, e2 = 0;//e1 e2为s1 s2的指数
s3 = deal(s1, e1);
s4 = deal(s2, e2);
if (s3 == s4 && e1 == e2) {//主体相同且指数相同
cout<<"YES 0."<< s3 <<"*10^"<< e1 <<endl;
}else{
cout<<"NO 0."<< s3 <<"*10^"<< e1<<" 0."<<s4 <<"*10^"<< e2 <<endl;
}
return 0;
}

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