[天坑]1060 Are They Equal （25 分)

2019-03-30

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 0.123×105with 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.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:

1	3 12300 12358.9

Sample Output 1:

1	YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

1	NO 0.12010^3 0.12810^3

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出有效位数和两个数字，要求比较这两数字用科学技术法表示后是否相同。例如题目所给3 12300 12358.9，都可以表示成0.123*10^5。

分析

这道题目简直就是天坑，巨坑。判断两个数相同，需要判断系数，指数都相同。使用string存储数。

对于系数，遍历，找到第一个不为0且不为.的数字，从这个数字开始取n个数。注意需要把小数点去掉，比如1.23，n=3，那么从1开始，取123三个数。

对于指数，找到第一个不为0且不为小数点的位置nonzero，找到小数点位置dot。dot-nonzero即为指数。例如1234.5。dot=4,nonzero=0。指数为4。指数小于0时需加一。例如0.0123，dot=1，nonzero=3，指数为1-3+1=-1。

得到系数指数，判断，输出即可。还需要注意输入为0的情况，手动在0.后面补充n个0。对应最后一个测试点。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;

void standform(string s,string r,int n){
    int dots=-1,nonzeros=-1,dotr=-1,nonzeror=-1;
    int c=0;
    string ns,nr;
    for(int i=0;i<s.length();i++){
        //第一个不为0不为.的位置
        if(s[i]!='.'&&s[i]!='0'&&nonzeros==-1)
            nonzeros = i;
        //从这个位置开始取n个数
        if(nonzeros!=-1&&c<n){
            if(s[i]!='.'){
                ns+=s[i];
                c++;
            }
        }
        //小数点位置
        if(s[i]=='.')
            dots=i;
    }
    if(dots==-1)
        dots=s.length();
    c=0;
    //同样处理
    for(int i=0;i<r.length();i++){
        if(r[i]!='.'&&r[i]!='0'&&nonzeror==-1)
            nonzeror = i;

        if(nonzeror!=-1&&c<n){
            if(r[i]!='.') {
                nr += r[i];
                c++;
            }
        }
        if(r[i]=='.')
            dotr=i;
    }
    if(dotr==-1)
        dotr=r.length();
    int exps=dots-nonzeros,expo=dotr-nonzeror;
    exps=exps<0?exps+1:exps;
    expo=expo<0?expo+1:expo;
    //处理输入为0情况
    if(nonzeros==-1) {
        ns = string(n, '0');
        exps=0;
    }
    if(nonzeror==-1) {
        nr = string(n, '0');
        expo=0;
    }
    //输出
    if(ns==nr&&exps==expo)
        cout<<"YES 0." + ns + "*10^" + to_string(exps);
    else
    {
        cout<<"NO";
        cout<<" 0." + ns + "*10^" + to_string(exps);
        cout<<" 0." + nr + "*10^" + to_string(expo);
    }
}
int main() {
    int n;
    string a,b;
    cin>>n>>a>>b;
    standform(a,b,n);
    return 0;
}