1060. Are they equal (25)-pat

1060. Are they equal (25) Time Limit 50 ms memory limit 32000 kb code length limit 16000 B discriminant program standard author Chen, Yue

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 A and 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 .. 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

Recommendation index :※※

Source: http://pat.zju.edu.cn/contests/pat-a-practise/1060

This question does not have a complex algorithm. You only need to consider several cases.

1. There is 0 in front of the number, for example, 5 003 03

2. Zero case, such as 3 0.0000 00.0

3. The calculation of indexes should be well considered.

#include<iostream>#include<string.h>using namespace std;#define  N 200int formatstr(char *str,char *sa,int len,int strsize){int i;int exp_num=0;int isbig=0,first_use=0;//first_use is the first use number,because of the case may have such 003 00.03for(i=0;i<strsize;i++){//check num is large than 1 or notif(str[i]>='1'&&str[i]<='9'){first_use=i;isbig=1;break;}else if(str[i]=='.'){first_use=i;isbig=0;break;}}int t,is_point_done,has_num;if(isbig==1){// num is large than 1is_point_done=0;for(i=first_use,t=0;t<len;i++){if(i<strsize){if(str[i]>='0'&&str[i]<='9'){sa[t++]=str[i];if(0==is_point_done)exp_num++;}else if(str[i]=='.')is_point_done=1;}elsesa[t++]='0';}for(;i<strsize&&is_point_done==0;i++){//check number whether arrive point '.' if(str[i]=='.')break;exp_num++;}}else{has_num=0;for(i=first_use+1;i<strsize&&str[i]=='0';i++)//skip 0exp_num--;for(t=0;t<len;i++){if(i<strsize){sa[t++]=str[i];if(str[i]>='1'&&str[i]<='9')has_num=1;}elsesa[t++]='0';}if(0==has_num)//check whether the case of 00.000000exp_num=0;}sa[t]='\0';return exp_num;}void print_num(char *a,int len,int exp_num){int i;cout<<"0.";for(i=0;i<len;i++)cout<<a[i];cout<<"*10^"<<exp_num;}int main(){int i,n,exp_num_a,exp_num_b;char a[N],b[N];char sa[N],sb[N];cin>>n>>a>>b;exp_num_a=formatstr(a,sa,n,strlen(a));//format for two numberexp_num_b=formatstr(b,sb,n,strlen(b));if(exp_num_a==exp_num_b){for(i=0;i<n;i++){if(sa[i]!=sb[i]){cout<<"NO ";print_num(sa,n,exp_num_a);cout<<" ";print_num(sb,n,exp_num_b);break;}}if(i==n){cout<<"YES ";print_num(sa,n,exp_num_a);}}else{cout<<"NO ";print_num(sa,n,exp_num_a);cout<<" ";print_num(sb,n,exp_num_b);}return 0;}