1060. Are They Equal (25)
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
code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;
struct node{
char d[105];
int e;
};
node getResult(char s[],int n){
int firstp = -1;
int pointp = -1;
int index = 0;
int i;
node r;
for(i = 0; s[i]; i++){
if(s[i] == '.'){
pointp = i;
continue;
}
else if(s[i] == '0' && firstp == -1){
continue;
}
else{
if(firstp == -1) firstp = i;
if(index < n){
if(index < strlen(s))
r.d[index++] = s[i];
else
r.d[index++] = '0';
}
}
}
while(index > 0 && index < n){
r.d[index++] = '0';
}
r.d[index] = 0;
if(pointp == -1) pointp = i;
if(pointp - firstp < 0)
r.e = pointp - firstp + 1;
else
r.e = pointp - firstp;
if(index == 0){
for(int i = 0; i < n; i++){
r.d[i] = '0';
}
r.d[n] = 0;
r.e = 0;
}
return r;
}
int main(){
int digit;
char a[1000],b[1000];
scanf("%d %s %s",&digit,a,b);
node s1 = getResult(a,digit);
node s2 = getResult(b,digit);
if(strcmp(s1.d,s2.d) == 0 && s1.e == s2.e){
printf("YES 0.%s*10^%d\n",s1.d,s1.e);
}
else{
printf("NO 0.%s*10^%d 0.%s*10^%d\n",s1.d,s1.e,s2.d,s2.e);
}
return 0;
}