PAT 甲级 1060 Are They Equal (25 分)(科学计数法,接连做了2天,考虑要全面,坑点多,真麻烦)

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 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 (<) 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 1, 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

 

题意:

给出两个数,问:将他们写成保留N (<100)位小数的科学计数法 0.d[1]...d[N]*10^k (d[1]>0 unless the number is 0); 后,是否相同。若相等,输出YES,并转换结果; 如果不相等,输出NO,并给出两个数的转换结果。

题解:

本题的思路难想而且情况判断非常复杂。

题目要求科学计数法时,两个数是否相等。因此只要判断科学技术法时的本体部分以及指数部分是否相等即可。

对于数据来说,要分为大于 1 与小于 1 来判断,要考虑各种情况下的数字,比如:0000, 000.00, 00123.4, 0.012, 0.00 等

一开始没有考虑太多的异常情况,拿了19分,第二天看了题解的注意点重做,还是只有21分,

第三天理了理思路,其实,只要关注,格式化好后的小数点要点在哪里,次数是多少就可以了,在此基础上,特判0,去掉前导零。比较的时候注意不要超限,输出不够0补齐。

 

自己编的测试样例:

2 0.0015 0000.001520000
YES 0.15*10^-2
3 010.25 0010.23
YES 0.102*10^2
5 0.1 00.100
YES 0.10000*10^0
10 000.0012 0000000.0012000000000
YES 0.1200000000*10^-2
3 0 0.000
YES 0.000*10^0

 

AC代码:

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int n;
char a[105];
char b[105];
int main(){
    cin>>n>>a>>b;
    int la=strlen(a);
    int lb=strlen(b);
    int sta=-1,stb=-1;
    int isa0=0,isb0=0;//特判是不是0

    //从第一个不是0的数开始
    for(int i=0;i<la;i++){
        if(a[i]!='.'&&a[i]!='0'){
            sta=i;
            break;
        }
    }
    for(int i=0;i<lb;i++){
        if(b[i]!='.'&&b[i]!='0'){
            stb=i;
            break;
        }
    }
    if(sta==-1) isa0=1;
    if(stb==-1) isb0=1;

    //小数点前有几位
    int ia=0,ib=0;
    int f=0;
    for(int i=0;i<la;i++){
        if(a[i]=='.') break;
        if(a[i]!='0') f=1;
        if(f){
            ia++;
        }
    }
    f=0;
    for(int i=0;i<lb;i++){
        if(b[i]=='.') break;
        if(b[i]!='0') f=1;
        if(f){
            ib++;
        }
    }
    //cout<<"sta:"<<sta<<" ia:"<<ia<<" ha:"<<ha<<endl;
    //cout<<"stb:"<<stb<<" ib:"<<ib<<" hb:"<<hb<<endl;

    //比较
    f=1;
    int pa=sta,pb=stb;
    if(isa0!=isb0 || ia!=ib) f=0;
    for(int i=0;i<n;i++){
        if(pa+i>=la||pb+i>=lb) break;
        if(a[pa]=='.') pa++;
        if(b[pb]=='.') pb++;
        //cout<<"pa "<<pa+i<<" a[pa] "<<a[pa+i]<<endl;
        //cout<<"pb "<<pa+i<<" b[pb] "<<b[pa+i]<<endl;
        if(a[pa+i]!=b[pb+i]){
            f=0;
            break;
        }
    }
    if(isa0==1&&isa0==isb0) f=1;//如果两个都是0

    //输出
    if(f){
        cout<<"YES 0.";
        if(isa0==1){//0的特判
            for(int i=1;i<=n;i++) cout<<"0";
            cout<<"*10^0";
        }else{
            pa=sta;
            for(int i=0;i<n;i++){
                if(pa+i>=la) {
                    cout<<"0";
                    continue;
                }
                if(a[pa+i]=='.') pa++;
                if(pa+i>=la) cout<<"0";
                else cout<<a[pa+i];
            }
            cout<<"*10^";
            if(ia!=0) cout<<ia;//次数>=于0
            else{//次数<0
                int k=-1;
                for(int i=0;i<la;i++){
                    if(a[i]=='.'){
                        k=i;
                        break;
                    }
                }
                cout<<k-sta+1;//.的位置-开始的位置+1
            }
        }
    }else{//同上
        cout<<"NO 0.";
        if(isa0==1){
            for(int i=1;i<=n;i++) cout<<"0";
            cout<<"*10^0 ";
        }else{
            for(int i=0;i<n;i++){
                if(pa+i>=la) {
                    cout<<"0";
                    continue;
                }
                if(a[pa+i]=='.') pa++;
                if(pa+i>=la) cout<<"0";
                else cout<<a[pa+i];
            }
            cout<<"*10^";
            if(ia!=0) cout<<ia<<" 0.";
            else{
                int k=-1;
                for(int i=0;i<la;i++){
                    if(a[i]=='.'){
                        k=i;
                        break;
                    }
                }
                cout<<k-sta+1<<" 0.";
            }
        }
        if(isb0==1){
            for(int i=1;i<=n;i++) cout<<"0";
            cout<<"*10^0";
        }else{
            for(int i=0;i<n;i++){
                if(pb+i>=lb) {
                    cout<<"0";
                    continue;
                }
                if(b[pb+i]=='.') pb++;
                if(pb+i>=lb) cout<<"0";
                else cout<<b[pb+i];
            }
            cout<<"*10^";
            if(ib!=0) cout<<ib;
            else{
                int k=-1;
                for(int i=0;i<lb;i++){
                    if(b[i]=='.'){
                        k=i;
                        break;
                    }
                }
                cout<<k-stb+1;
            }
        }
    }
    return 0;
}

 

posted on 2019-09-23 14:32  蔡军帅  阅读(496)  评论(0编辑  收藏  举报