1060. Are They Equal (25)

红太狼 2022-05-29 23:00 137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;
    }

参考： [柳婼][Link 1]

[Link 1]: http://blog.csdn.net/liuchuo/article/details/52497386