1060. Are They Equal (25)-蒲公英云

1060. Are They Equal (25)

红太狼 2022-05-29 23:00 238阅读 0赞

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…dN*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

题目大意：

代码：

#include<stdio.h>
#include<string.h>
char A[1000],B[1000],a[1000],b[1000];
int main()
{
    int i,j,n,cnta,cntb,pa,pb,indexa,indexb;
    scanf("%d %s %s",&n,a,b);
    cnta=strlen(a);
    cntb=strlen(b);
    for(i=0;i<strlen(a);i++)
    {
        if(a[i]=='.')
        {
            cnta=i;
            break;
        }
    }
    for(i=0;i<strlen(b);i++)
    {
        if(b[i]=='.')
        {
            cntb=i;
            break;
        }
    }
    pa=0;
    while(a[pa]=='0'||a[pa]=='.')
        pa++;
        pb=0;
    while(b[pb]=='0'||b[pb]=='.')
        pb++;
    if(cnta>=pa)
    {
        cnta=cnta-pa;
    }
    else
    {
        cnta=cnta-pa+1;
    }
    if(cntb>=pb)
    {
        cntb=cntb-pb;
    }
    else
    {
        cntb=cntb-pb+1;
    }
    if(pa==strlen(a))
        cnta=0;
    if(pb==strlen(b))
        cntb=0;
    indexa=0;
    while(indexa<n)
    {
        if(a[pa]!='.'&&pa<strlen(a))
        {
            A[indexa++]=a[pa];
        }
        else if(pa>=strlen(a))
        {
            A[indexa++]='0';
        }
        pa++;
    }
    A[indexa]='\0';
    indexb=0;
    while(indexb<n)
    {
        if(b[pb]!='.'&&pb<strlen(b))
        {
            B[indexb++]=b[pb];
        }
        else if(pb>=strlen(b))
        {
            B[indexb++]='0';
        }
        pb++;
    }
    B[indexb]='\0';
    if(strcmp(A,B)==0&&cnta==cntb)
    {
        printf("YES 0.%s*10^%d\n",A,cnta);
    }
    else
    {
        printf("NO 0.%s*10^%d 0.%s*10^%d\n",A,cnta,B,cntb);
    }
    return 0;
}