模拟_1060 Are They Equal （25 分）

1060 Are They Equal （25 分）

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 0.123×10​5​​ 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 Aand B are the two float numbers to be compared. Each float number is non-negative, no greater than 10​100​​, 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.d[1]...d[N]*10^k (d[1]>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

这个题有好多细节需要考虑 ，首先是前导零，例如000.5的情况

然后如果去除前导零之后，第一位为‘.’，这是要判断后面是否接着是0，例如000.0005    这时，需要继续向后删除，因为只有为0的情况下，d[1]才能为0

最后取出完前导零，判断字符串是否还有剩余，没有的话，就证明这个字符串为0，否则不为0

接着计算有效数字，计算偏移量即可

#include <vector>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 0x3f3f3f3f

using namespace std;
const int maxn = 105;

char * change(int N,string a)
{
    char *num = (char *)malloc(N*sizeof(char));
    int index = 0,len = a.length(),j = 0;
    for(int i = 0;i < N;i ++)
    {
        if(a[j] == '.')
        {
            j++;i--;continue;
        }
        if(j < len)
            num[index++] = a[j++];
        else
        {
            while(i++ < N)
                num[index++] = '0';
            break;
        }
    }
    return num;
}
void format(string &a,int &index)
{
    while(a[0] == '0') a.erase(0,1);    //前导零   
    while(a[0] == '.') a.erase(0,1);    //如果小数点后面也是零
    while(a[0] == '0') a.erase(0,1),index --;   //也要去除000.005
}
int main()
{
    int N;
    string a,b;
    cin>>N>>a>>b;
    int indexa=0,indexb = 0;

    format(a,indexa);
    format(b,indexb);

    int lena = a.length(),lenb = b.length();
    if(lena == 0) indexa = 0;
    if(lenb == 0) indexb = 0;

    if(indexa == 0)
        for(int i =0 ;i < lena;i ++) if(a[i] != '.') indexa ++ ;else break;
    if(indexb == 0)
        for(int i = 0;i < lenb;i ++) if(b[i] != '.') indexb ++ ;else break;

    char *aa = change(N,a);
    char *bb = change(N,b);
    if(strcmp(aa,bb) == 0 && indexa == indexb)
    {
        printf("YES ");
        cout << "0."<<aa<<"*10^"<<indexa<<endl;
    }
    else
    {
        printf("NO ");
        cout << "0."<<aa<<"*10^"<<indexa<< " 0."<<bb<<"*10^"<<indexb<<endl;
    }
}