Codeforces 600A Extract Numbers 【模拟】

本文介绍如何从给定的字符串中分离出数字和非数字部分,并将它们分别组成新的字符串。通过模拟方法实现这一过程,确保数字部分形成新字符串a,非数字部分形成新字符串b。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A. Extract Numbers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 strings).

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".

Input

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.

Output

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.

Sample test(s)
input
aba,123;1a;0
output
"123,0"
"aba,1a"
input
1;;01,a0,
output
"1"
",01,a0,"
input
1
output
"1"
-
input
a
output
-
"a"
Note

In the second example the string s contains five words: "1", "", "01", "a0", "".



好弱。。。WA三次才过。


题意:给定一个字符串,字符';' 和 ','将串分开,每部分是一个单词,若出现中间为空的情况(连续两个分隔符),则认为中间的单词是空格。若单词是整数(不能有前导0),则放在串a里面,反之放在串b里面。在a,b串中,单词用','分开。若无单词是整数,a串为"-",若无单词是非整数,b串为"-"。最后输出a串和b串,如a串(b串)不是"-",输出时用""括起。



先填上空格,直接模拟就可以了。


AC代码:


#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (100000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
char s[MAXN], str[MAXN*2];
char a[MAXN*2], b[MAXN*2];
bool judge(char op){
    return op != ',' && op != ';';
}
bool check(char op){
    return (op >= 'a' && op <= 'z') || (op >= 'A' && op <= 'Z') || (op == ' ') || (op == '.');
}
int main()
{
    while(scanf("%s", s) != EOF)
    {
        int len = strlen(s), l;
        if(len == 1 && (s[0] == ';' || s[0] == ','))
        {
            printf("-\n");
            printf("%c,%c\n", 34, 34);
            continue;
        }
        int top = 0;
        for(int i = 0; i < len; i++)
        {
            if(i == 0)
            {
                if(s[i] == ',' || s[i] == ';')
                {
                    str[top++] = ' ';
                    str[top++] = s[i];
                    if(s[i+1] == ',' || s[i+1] == ';')
                        str[top++] = ' ';
                }
                else
                    str[top++] = s[i];
            }
            else if(i == len-1)
            {
                if(s[i] == ',' || s[i] == ';')
                {
                    str[top++] = s[i];
                    str[top++] = ' ';
                }
                else
                    str[top++] = s[i];
            }
            else
            {
                if((s[i] == ',' || s[i] == ';') && (s[i+1] == ',' || s[i+1] == ';'))
                {
                    str[top++] = s[i];
                    str[top++] = ' ';
                }
                else
                    str[top++] = s[i];
            }
        }
        str[top] = '\0';
        a[0] = b[0]= '"'; a[1] = b[1] = '\0';
        int timea = 0, timeb = 0;
        for(int i = 0; i < top; )
        {
            if(judge(str[i]))
            {
                int j = i; int k = 0; bool flag = true;
                while(j < top && judge(str[j]))
                {
                    if(check(str[j]))
                        flag = false;
                    s[k++] = str[j++];
                }
                s[k] = '\0';
                if(s[0] == '0' && k > 1)
                    flag = false;
                if(flag)
                {
                    if(timea)
                    {
                        l = strlen(a);
                        a[l] = ',';
                        a[l+1] = '\0';
                    }
                    if(s[0] != ' ')
                        strcat(a, s);
                    timea++;
                }
                else
                {
                    if(timeb)
                    {
                        l = strlen(b);
                        b[l] = ',';
                        b[l+1] = '\0';
                    }
                    if(s[0] != ' ')
                        strcat(b, s);
                    timeb++;
                }
                i = j+1;
            }
        }
        if(timea == 0)
            printf("-\n");
        else
        {
            printf("%s", a);
            printf("%c\n", 34);
        }
        if(timeb == 0)
            printf("-\n");
        else
        {
            printf("%s", b);
            printf("%c\n", 34);
        }
    }
    return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值