头文件
#include<string>
using namespace std;
string str;
string str = "Gqq";
#include<cstdio>
#include<string>
using namespace std;
int main(){
string str = "gqq";
for(int i = 0; i < str.length(); i++){
printf("%c", str[i]);
}
return 0;
}
如果要读入和输出整个字符串，只能用cin和cout
#include<iostream>
#include<string>
using namespace std;
int main(){
string str;
cin >> str;
cout << str;
return 0;
}
用printf输出string时，用c_str()将string类型转换为字符数组进行输出
#include<iostream>
#include<string>
using namespace std;
int main(){
string str;
cin >> str;
printf("%s", str.c_str());
return 0;
}
定义迭代器
string::iterator it;
通过迭代器*it来访问string里的每一位
#include<iostream>
#include<string>
using namespace std;
int main(){
string str;
cin >> str;
for(string::iterator it = str.begin(); it != str.end(); it++){
printf("%c", *it);
}
return 0;
}
string和vector一样，支持直接对迭代器进行加减某个数字，如str.begin() + 3
string的加法， 将两个string直接拼接起来
#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "Gqq";
string str2 = " loves apples!";
string str3 = str1 + str2;
str1 += str2; // str2直接拼接在str1上
cout << str3 << endl;
cout << str1 << endl;
return 0;
}
Gqq loves apples!
Gqq loves apples!
两个string类型可以直接使用
==、!=、<、<=、>、>=比较大小，比较规则是字典序
#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "aa";
string str2 = "aaa";
string str3 = "abc";
string str4 = "gqq";
if(str1 < str2) printf("ok1\n");
if(str1 != str3) printf("ok2\n");
if(str4 >= str3) printf("ok3\n");
return 0;
}
length( ) 返回string的长度，即存放的字符数，时间复杂度为O(1)。size( )和length( )基本相同
string str = "gqq";
printf("%d" %d\n", str.length(), str.size());
3 3
#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "Gqq apples";
string str2 = " loves";
str1.insert(3, str2);
printf("%s", str1.c_str());
return 0;
}
Gqq loves apples
it1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，用来表示串[it2, it3)将被插在it1的位置上#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "Gqq apples";
string str2 = " loves";
str1.insert(str1.begin() + 3, str2.begin(), str2.end());
cout << str1 << endl;
return 0;
}
Gqq loves apples
str.erase(it)用于删除单个元素，it为需要删除的元素的迭代器
#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "Gqq loves apples";
str1.erase(str1.begin() + 2);
cout << str1 << endl;
return 0;
}
Gq loves apples
str.erase(first, last)，其中first为需要删除的区间的其实迭代器，而last则为需要删除区间的末尾迭代器的下一个地址，即[first, last)
#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "Gqq loves apples";
str1.erase(str1.begin() + 2, str1.begin() + 9);
cout << str1 << endl;
return 0;
}
Gq apples
str.erase(pos, length)，pos为需要开始删除的起始位置，length
#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "Gqq loves apples";
str1.erase(3, 6); // 删除从3号位开始的6个字符
cout << str1 << endl;
return 0;
}
Gqq apples
clear( ) 用以清空string中的数据，时间复杂度为O(1)
#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "Gqq loves apples";
str1.clear();
cout << str1.length() << endl;
return 0;
}
Result ： 0
substr(pos, len) 返回从pos号位开始，长度为len的子串，时间复杂度为O(len)
#include<iostream>
#include<string>
using namespace std;
int main(){
string str1 = "Gqq loves apples";
string str2 = str1.substr(0, 3);
cout << str2 << endl;
return 0;
}
Gqq
string::npos是一个常数，其本身的值为-1， 但是由于是unsigned_int类型，因此实际上也可以认为是unsigned_int类型的最大值。string::npos用以作为find函数失配时的返回值。
#include<iostream>
#include<string>
using namespace std;
int main(){
if(string::npos == -1)
cout << "-1 is true" << endl;
if(string::npos == 4294967295)
cout << "4294967295 is true" << endl;
return 0;
}
-1 is true
4294967295 is true
// 可以认为string::npos就是可以等于-1或者4294967295
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，返回string::npos
str.find(str2, pos)，从str的pos号位开始匹配str2，返回值与上面相同
时间复杂度为O(nm)，其中n和m分别是str和str2的长度
#include<iostream>
#include<string>
using namespace std;
int main(){
string str = "Thank you for your smile";
string str2 = "you";
string str3 = "me";
if(str.find(str2) != string::npos){
cout << str.find(str2) << endl;
}
if(str.find(str2, 7) != string::npos){
cout << str.find(str2, 7) << endl;
}
if(str.find(str3) != string::npos){
cout << str.find(str3) << endl;
}else{
cout << "I know there is no position for me." << endl;
}
return 0;
}
6
14
I know there is no position for me.
str.replace(pos, len, str2)把str从pos号位开始，长度为len的子串替换为str2
str.replace(it1, it2, str2)把str的迭代器[it1, it2)范围的子串替换为str2
#include<iostream>
#include<string>
using namespace std;
int main(){
string str = "Thank you for your smile";
string str2 = "the dinner";
string str3 = "Thanks";
cout << str.replace(14, 10, str2) << endl;
cout << str.replace(str.begin(), str.begin() + 9, str3);
return 0;
}
Thank you for the dinner
Thanks for the dinner
题目描述：
If a machine can save only 3 significant digits, the float numbers 12 300 and 12 358.9 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.
输入格式:
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.
输出格式:
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.d1…dN*10^k” (d1>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.
输入样例1:
3 12300 12358.9
输出样例1:
YES 0.123*10^5
输入样例2:
3 120 128
输出样例2:
NO 0.120*10^3 0.128*10^3
解题思路
我们可能遇到大于0或者小于0的两种情况数字，如
但是，由于可能隐含的出现数字前面有多个零，如
所以，对于求解，有两个分类：
通过上面步骤，即可得到指数e，之后根据精度将字符串拼接即可，代码如下：
#include<iostream>
#include<string>
using namespace std;
int n; // 有效位数
string deal(string s, int& e){
int k = 0; // S的下标
while(s.length() > 0 && s[0] == ‘0‘){
s.erase(s.begin()); // 去掉s的前导零
}
if(s[0] == ‘.‘){ // 去掉前导零后是小数点，说明s小于1
s.erase(s.begin()); // 去掉小数点
while(s.length() > 0 && s[0] == ‘0‘){
s.erase(s.begin());
e--;
}
} else { // 去掉前导零后不是小数点，则找到后面的小数点删除
while(k < s.length() && s[k] != ‘.‘){
k++;
e++;
}
}
if(k < s.length()){ // while结束后k < s.length()，说明碰到了小数点
s.erase(s.begin() + k); // 删除小数点
}
if(s.length() == 0){
e = 0; // 如果去除前导零后s的长度变为0，说明这个数是0
}
int num = 0;
k = 0;
string res;
while(num < n){ // 只要精度还没有到n
if(k < s.length()) res += s[k++]; // 只要还有数字，就加到res末尾
else res += ‘0‘; // 否则res末尾添加0
num++; // 精度加1
}
return res;
}
int main(){
string s1, s2, s3, s4;
cin >> n >> s1 >> s2;
int e1 = 0, e2 = 0;
s3 = deal(s1, e1);
s4 = deal(s2, e2);
if(s3 == s4 && e1 == e2)
cout << "YES 0." << s3 << "*10^" << e1 << endl;
else
cout << "NO 0." << s3 << "*10^" << e1 << " 0." << s4 << "*10^" << e2 << endl;
return 0;
}
结果
3 12300 12358.9
YES 0.123*10^5
Write by Gqq
原文：https://www.cnblogs.com/zgqcn/p/12584556.html