PAT 1060 Are They Equal[难][科学记数法]

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 0.123×105 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.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

题目大意：机器只能存储N位数，给出一个最多100位的浮点数，判断它们在机器中是否是相同的存储。

//一开始想用字符数组，但是在字符数组中查找小数点的位置还要遍历，不如使用string直接find方便。

#include <iostream>

#include<stdio.h>

#include<cmath>

#include<vector>

using namespace std;

int main() {

    int n;

    string s1,s2;

    cin>>n;

    cin>>s1>>s2;

    int len1,len2;

    int pos=s1.find(".");

    pos==string::npos?len1=s1.size():len1=pos;

    pos=s2.find(".");

    pos==string::npos?len2=s2.size():len2=pos;

    string temp1=s1.substr(,n);

    string temp2=s2.substr(,n);

    if(len1!=len2){

        cout<<"NO ";

        cout<<"0."<<temp1<<"*10^"<<len1<<" ";

        cout<<"0."<<temp2<<"*10^"<<len2;

    }else{

        if(temp1==temp2){

            cout<<"YES ";

            cout<<"0."<<temp1<<"*10^"<<len1;

        }

        else{

        cout<<"NO ";

        cout<<"0."<<temp1<<"*10^"<<len1<<" ";

        cout<<"0."<<temp2<<"*10^"<<len2;

        }

    }

    return ;

}

//提交一次之后，没有通过，牛客网上报错：

测试用例:
99 0 0.000

对应输出应该为:

YES 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*10^0

你的输出为:

NO 0.0*10^1 0.0.000*10^1

//但是我发现我不知道怎么处理，我写的这个bug存在问题，

//看了大佬的代码之后发现，我对科学计数法并不太明白。

代码来自：https://www.liuchuo.net/archives/2293

#include <iostream>

#include <cstring>

#include<stdio.h>

using namespace std;

int main() {

    int n, p = , q = ;

    char a[], b[], A[], B[];

    scanf("%d%s%s", &n, a, b);

    int cnta = strlen(a), cntb = strlen(b);

    for(int i = ; i < strlen(a); i++) {//居然真的是遍历，我还以为需要字符串那样呢。

        if(a[i] == '.') {

            cnta = i;

            break;

        }

    }

    for(int i = ; i < strlen(b); i++) {

        if(b[i] == '.') {

            cntb = i;

            break;

        }

    }

    while(a[p] == '' || a[p] == '.') p++;//找到不为0的位置。

    while(b[q] == '' || b[q] == '.') q++;

    if(cnta >= p)//这个小数点出现在非0位之后，如123.45，指数就为cnta-p=3

        cnta = cnta - p;

    else//小数点后仍有0位，如0.0002，那么指数就是负值，

        cnta = cnta - p + ;

    if(cntb >= q)

        cntb = cntb - q;

    else

        cntb = cntb - q + ;//

    if(p == strlen(a))//如果是0.000这样的形式，将其赋值为0.

        cnta = ;//不置为0，有可能两个0判断不相等。

    if(q == strlen(b))

        cntb = ;

    int indexa = , indexb = ;

    while(indexa < n) {

        if(a[p] != '.' && p < strlen(a))//除了小数点以外都赋值，

            A[indexa++] = a[p];

        else if(p >= strlen(a))//当遍历完后，n较大时，仍然不够n位，那么后n位补0

            A[indexa++] = '';

        p++;

    }

    while(indexb < n) {

        if(b[q] != '.' && q < strlen(b))

            B[indexb++] = b[q];

        else if(q >= strlen(b))

            B[indexb++] = '';

        q++;

    }

    if(strcmp(A, B) ==  && cnta == cntb)

        printf("YES 0.%s*10^%d", A, cnta);//读入和输出字符数组都可以用%s.

    else

        printf("NO 0.%s*10^%d 0.%s*10^%d" , A, cnta, B, cntb);

    return ;

}

//深刻地学习了科学记数法

1.科学记数法有两个重要的因素，就是小数点的位置，和首位非0位的位置，通过这两个可以确定指数，

2.小数点的位置，怎么办呢？先初始化位字符串长度，对应整数，没有小数位。

3.首位非0的元素也初始化为字符串的长度。

4.需要注意的是如果首位非零元长度=字符串长度，那么这个字符串表示的数就是0.需要特殊处理，就是为0.0..0*10^0。否则计算指数。

5.科学记数法如果不是0，那么处理后的结果，小数点后边一定要是非0元！。

学习了！

巴特西