PAT A1060——string的常见用法详解

string 常用函数实例

(1)operator += 可以将两个string直接拼接起来

(2)compare operator 可以直接使用==、!=、<、<=、>、>=比较大小，比较规则是字典序

(3)length()/size()

(4)clear():清空所有元素

(5)erase():erase(st.begin()+3)删除第四个元素；erase(first,last)删除【first,last)内所有元素；erase(pos,length)pos为需要开始删除的起始位置，length为删除的字符个数

(6)insert():insert(pos,string)在pos位置插入字符串;insert（it,it2,it3)it为原字符串的欲插入位置，it2和it3为待插字符串的首尾迭代器

(7)substr:substr(pos,len)返回从pos号位开始，长度为len的子串

(8)string::npos:是一个常数，等于-1或者4294967295，用以作为find函数失配时的返回值

(9)find():str.find(str2),当str2是str的子串时，返回其在str中第一次出现的位置，如果str2不是str的子串则返回string::npos

(10)replace():str.replace(pos,len,str2)把str从pos号位开始，长度为len的子串替换为str2

 

1060 Are They Equal

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]...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位小数的科学计数法后是否相等。如果想等，则输出“YES”，并给出该转换结果；如果不相等，则输出“NO”，并分别给出两个数的转换结果

参考代码：

 1 #include<iostream>
 2 #include<string>
 3 using namespace std;
 4 int n;  //有效位数
 5 string deal(string s, int& e) {
 6     int k = 0;  //s的下标
 7     while(s.length() > 0 && s[0] == '0') {
 8         s.erase(s.begin());     //去掉s的前导零
 9     }
10     if(s[0] == '.') {       //若去掉前导零后是小数点，则说明s是小于1的小数
11         s.erase(s.begin()); //去掉小数点
12         while(s.length() > 0 && s[0] == '0') {
13             s.erase(s.begin());     //去掉小数点后非零位前的所有零
14             e--;        //每去掉一个0，指数e减一
15         }
16     }
17     else{       //若去掉前导零后不是小数点，则找到后面的小数点删除
18         while(k<s.length() && s[k] != '.') {    //寻找小数点
19             k++;
20             e++;    //只要不遇到小数点，就让指数e++
21         }
22         if(k < s.length()) {    //while结束后k < s.length(),说明遇到了小数点
23             s.erase(s.begin() + k);     //把小数点删除
24         }
25     }
26     if(s.length() == 0) {
27         e = 0;      //如果去除前导零后s的长度变为0,则说明这个数是0
28     }
29     int num = 0;
30     k = 0;
31     string res;
32     while(num < n) {    //只要精度还没有到n
33         if(k < s.length()){
34             res += s[k++];  //只要还有数字，就加到res末尾
35         }
36         else{
37             res += '0';     //否则res末尾添加0
38         }
39         num++;
40     }
41     return res;
42 }
43
44 int main() {
45     string s1, s2, s3, s4;
46     cin >> n >> s1 >> s2;
47     int e1 = 0,e2 = 0;  //e1,e2为s1与s2的指数
48     s3 = deal(s1,e1);
49     s4 = deal(s2,e2);
50     if(s3 == s4 && e1 == e2) {  //若主体相同且指数相同，则输出“YES”
51         cout<<"YES 0."<<s3<<"*10^"<<e1<<endl;
52     }
53     else{
54         cout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2<<endl;
55     }
56     return 0;
57 }