算法笔记 C++ STL学习（string的常见用法）

string：存放字符串
字符串的访问：
单个字符：字符数组访问方式
整个字符串：只能用cin，cout

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<stdlib.h>
    #include<string>
    using namespace std;
    int main(void){
        string str = "fdsfdsf";
        int i;
        for(i=0;i<str.length();i++){
            printf("%c",str[i]);
        }
        printf("\n");
        cout<<str;
        return 0;
    }

用printf输出string：使用c_str()将string转化为字符数组

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<stdlib.h>
    #include<string>
    using namespace std;
    int main(void){
        string str = "fdsfdsf";
        printf("%s",str.c_str());
    }

通过迭代器访问，与其他迭代器用法相同

	#include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<stdlib.h>
    #include<string>
    using namespace std;
	int main(void){
        string str = "fdsfdsf";
        string::iterator it;
    }

常用函数：
字符串拼接

    #include<stdio.h>
    #include<iostream>
    #include<string>
    using namespace std;
    int main(void){
        string str1 = "abc",str2 = "def",str3;
        str3 = str1 + str2;
        str1 += str2;
        cout<<str1;
        cout<<"\n";
        cout<<str3;
    }

字符串比较大小：（==，>=，<）
字符串的长度：size()，length()
字符串插入：
insert(pos,string)：在pos位置插入string

    #include<stdio.h>
    #include<iostream>
    #include<string>
    using namespace std;
    int main(void){
        string str1 = "abc",str2 = "def",str3;
        str1.insert(2,str2);
        cout<<str1;
    }

insert(it,it2,it3)：it为字符串的欲插入位置，it2和it3为待插字符串的首尾迭代器

    #include<stdio.h>
    #include<iostream>
    #include<string>
    using namespace std;
    int main(void){
        string str1 = "abc",str2 = "def",str3;
        str1.insert(str1.begin()+2,str2.begin(),str2.end());
        cout<<str1;
    }

erase()：删除元素
clear()：清空
substr()：返回从pos位开始，长度为len的子串

    #include<stdio.h>
    #include<iostream>
    #include<string>
    using namespace std;
    int main(void){
        string str1 = "abcdef",str2 = "def",str3;
        cout<<str1.substr(2,3);
    }

常数：string::npos find函数失配时的返回值，值为 -1 或4294967295
find(str)：返回子串第一次出现的位置
find(str,pos)：从pos位开始查找，返回子串第一次出现的位置

    #include<stdio.h>
    #include<iostream>
    #include<string>
    using namespace std;
    int main(void){
        string str1 = "Thank you for your smile";
        string str2 = "you";
        string str3 = "me";
        cout<<str1.find(str2);
        cout<<str1.find(str3);
        return 0;
    }

replace(pos,len,str2)：从pos位开始长度为len的字符串替换为str2
replace(it1,it2,str2)：[it1,it2)范围内的字符串替换为str2

    #include<stdio.h>
    #include<iostream>
    #include<string>
    using namespace std;
    int main(void){
        string str1 = "Thank you for your smile";
        string str2 = "you";
        string str3 = "me";
        str1.replace(2,4,str2);
        str1.replace(str1.begin(),str1.end(),str2);
        cout<<str1<<endl;
        return 0;
    }

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 0.123×10​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 YESif the two numbers are treated equal, and then the number in the standard form 0.d[1]...d[N]*10^k(d[1]>0unless the number is 0); or NOif 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

Code:

#include<stdio.h>
#include<iostream>
#include<string>
using namespace std;
int n;
string deal(string s,int &e){
    int k = 0;
    while(s.length() > 0 && s[0] == '0'){
        s.erase(s.begin());
    }
    if(s[0] == '.'){
        s.erase(s.begin());
        while(s.length() > 0 && s[0] == '0'){
            s.erase(s.begin());
            e--;
        }
    }
    else{
        while(k<s.length() && s[k] != '.'){
            k++;
            e++;
        }
        if(k < s.length()){
            s.erase(s.begin() + k);
        }
    }
    if(s.length() == 0){
        e = 0;
    }
    int num = 0;
    k = 0;
    string res;
    while(num < n){
        if(k < s.length()){
            res += s[k++];
        }
        else{
            res += '0';
        }
        num++;
    }
    return res;
}
int main(void){
    string str1,str2,str3,str4;
    cin>>n>>str1>>str2;
    int e1 = 0;
    int e2 = 0;
    str3 = deal(str1,e1);
    str4 = deal(str2,e2);
    if(str3 == str4 && e1 == e2){
        cout<<"YES 0."<<str3<<"*10^"<<e1<<endl;
    }
    else{
        cout<<"NO 0."<<str3<<"*10^"<<e1<<" 0."<<str4<<"*10^"<<e2;
    }
    return 0;
}