1060. Are They Equal (25)

1060. Are They Equal (25)

时间限制

100 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

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;
}