1060. Are They Equal (25)

题目连接：https://www.patest.cn/contests/pat-a-practise/1060

原题如下：

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

失落……
考察科学计数法，关键在于小数点和第一个有效位的位置的确定。
参考了他人的代码http://blog.csdn.net/xtzmm1215/article/details/38809629

#include<stdio.h>
#define MAXN 105
#include<string.h>
#include<stdlib.h>
struct result{
    char d[MAXN];
    int k;
};

result getResult(char *a,int n)
{
    int firstPos=-1,pointPos=-1;
    int index=0;
    result r;
    int i=0;
    for ( i=0;a[i];i++)
    {
        if (a[i]=='.')
        {
            pointPos=i;
            continue;
        }
        if (firstPos==-1 && a[i]=='0')continue;
        else
        {
            if (firstPos==-1 && a[i]!='0')
            {
                firstPos=i;
            }
            if (index<n)
            {
                r.d[index++]=a[i];
            }
        }
    }
        r.d[index]=0;

        if (pointPos==-1)pointPos=i;
        if (pointPos-firstPos<0)r.k=pointPos-firstPos+1;
        else r.k=pointPos-firstPos;

        int j;
        if (index==0)
        {
            for (j=0;j<n;j++)r.d[j]='0';
            r.d[j]=0;
            r.k=0;
        }
    return r;
}

int main()
{
    int N;
    scanf("%d",&N);
    char s1[MAXN],s2[MAXN];
    result r1,r2;
    scanf("%s %s",s1,s2);
    r1=getResult(s1,N);
    r2=getResult(s2,N);
    if (strcmp(r1.d,r2.d)==0 && r1.k==r2.k)
    {
        printf("YES 0.%s*10^%d",r1.d,r1.k);
    }
    else printf("NO 0.%s*10^%d 0.%s*10^%d",r1.d,r1.k,r2.d,r2.k);
    return 0;
}