You are given string s. Let's call word any largest sequence of consecutive symbols without symbols ',' (comma) and ';' (semicolon). For example, there are four words in string "aba,123;1a;0": "aba", "123", "1a", "0". A word can be empty: for example, the strings=";;" contains three empty words separated by ';'.
You should find all words in the given string that are nonnegative INTEGER numbers without leading zeroes and build by them new string a. String a should contain all words that are numbers separating them by ',' (the order of numbers should remain the same as in the string s). By all other words you should build string b in the same way (the order of numbers should remain the same as in the string s).
Here strings "101", "0" are INTEGER numbers, but "01" and "1.0" are not.
For example, for the string aba,123;1a;0 the string a would be equal to "123,0" and string b would be equal to "aba,1a".
The only line of input contains the string s (1 ≤ |s| ≤ 105). The string contains only symbols '.' (ASCII 46), ',' (ASCII 44), ';' (ASCII 59), digits, lowercase and uppercase latin letters.
Print the string a to the first line and string b to the second line. Each string should be surrounded by quotes (ASCII 34).
If there are no words that are numbers print dash (ASCII 45) on the first line. If all words are numbers print dash on the second line.
aba,123;1a;0
"123,0" "aba,1a"
1;;01,a0,
"1" ",01,a0,"
1
"1" -
a
- "a"
In the second example the string s contains five words: "1", "", "01", "a0", "".
题意:给一个字符串,里面的小字符串用','或者';'分开,筛选出所有的不含前导0的数字存在一个字符串里面,并用','分开,其余字符串存在另一个字符串里面,空字符串也要存,存在第二个字符串里面。
模拟题,不用string感觉很复杂的,难得处理,自己并不能做出来,看整场第二的代码,写的好简洁,处理得好好。遂学了一发。
把所有字符串处理出来,对每个字符串进行判断,归类到两种字符串中。对应输出就行
学到一个函数,substr(),筛选出指定字符串中指定两个位置之间的字符。
CODE
#include"stdio.h"
#include"algorithm"
#include"iostream"
#include"string.h"
#include"math.h"
#include"stdlib.h"
using namespace std;
string s;
vector <string> vs;
int check(char c)
{
if(c == ';' || c == ',')
return 1;
return 0;
}
int checknum(string vs) ///检查数字
{
if(!vs.size()) return 0;
if(vs.size() != 1 && vs[0] == '0') return 0;
for(int i = 0;i < (int)vs.size();i++)
{
if(vs[i] < '0' || vs[i] > '9')
return 0;
}
return 1;
}
int main(void)
{
cin>>s;
s += ","; ///加一个","方便处理最右的数据
int len = s.size();
int pos = 0;
for(int i = 0;i < len;i++)
{
if(check(s[i]))
{
vs.push_back(s.substr(pos,i-pos)); ///get新技能
pos = i+1;
}
}
string a,b;
for(int i = 0;i < (int)vs.size();i++)
{
if(checknum(vs[i]))
a += ","+vs[i];
else
b += ","+vs[i];
}
if(a.size())
{
a[0] = '"'; ///第一个为',' 改存为'"'
cout<<a<<"\""<<endl; ///'"'输出前面要加'\'
}
else cout <<"-"<<endl;
if(b.size())
{
b[0] = '"';
cout<<b<<"\""<<endl;
}
else cout<<"-"<<endl;
return 0;
}