1060. Are They Equal (25)

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本文介绍了一个程序设计问题：如何判断两个浮点数在一个仅保存特定数量有效数字的计算机系统中是否被视为相等。输入包括有效数字的数量及两个待比较的非负浮点数，输出则表明这两个数是否等价，并给出它们的标准形式。

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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;
}