1060. Are They Equal (25)

Posted 2020-10-06 给杰瑞一块奶酪~

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


一道让人心烦的题，需要注意这么几点，题目输出的有效数字位数不超过n；要确定第一个不为0的数的位置，以及小数点的位置，从而判断是几次方；再就是如果都是0，不存在不为0的数字，那么要记得补0，直到够了n位数。
小数点的位置减去第一个不为0的数的位置就是要求的指数，如果指数为负需要加1，比如a为0.002，在字符串中point_a=1,non0_a=4,1-4=-3,a可以表示为0.2*10^-2。



代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
void print(char *p,int k)///输出格式
{
    printf("0.%s*10^%d",p,k);
}
int main()
{
    int n,flag;
    char a[100],b[100],aa[100],bb[100];
    cin>>n>>a>>b;
    int non0_a = 0,non0_b = 0,point_a = strlen(a),point_b = strlen(b),count_a = 0,count_b = 0;
    int size_a = strlen(a),size_b = strlen(b);
    int index_a,index_b;
    flag=0;
    for(int i = 0;i < size_a;i++)
    {
        if(a[i] == ‘.‘)point_a = i;
        else if(a[i] != ‘0‘ || flag)
        {
            if(!flag)non0_a = i,flag = 1;
            if(count_a < n)aa[count_a++] = a[i];
        }
    }
    if(!count_a)
    {
        while(count_a<n)aa[count_a++]=‘0‘;
        point_a = 0;
    }
    aa[count_a] = ‘\0‘;
    index_a = point_a - non0_a;
    if(index_a < 0)index_a += 1;
    flag = 0;
    for(int i = 0;i < size_b;i++)
    {
        if(b[i] == ‘.‘)point_b = i;
        else if(b[i] != ‘0‘ || flag)
        {
            if(!flag)non0_b = i,flag = 1;
            if(count_b < n)bb[count_b++] = b[i];
        }
    }
    if(!count_b)
    {
        while(count_b<n)bb[count_b++]=‘0‘;
        point_b = 0;
    }
    bb[count_b] = ‘\0‘;
    index_b = point_b - non0_b;
    if(index_b < 0)index_b += 1;
    if(index_a == index_b && strcmp(aa,bb)==0)
    {
        cout<<"YES ";
        print(aa,index_a);
    }
    else
    {
        cout<<"NO ";
        print(aa,index_a);
        cout<<‘ ‘;
        print(bb,index_b);
    }
}