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 }