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题解P1321[单词覆盖还原]

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题目来源P1321

一共只会出现以下字符：

b o y g i r l .

把boy和girl两串中可能出现的重叠形式全部替换成某个字符，例如我用a..替换boy的组合，用z...替换girl的组合

得出boy和girl可能的拆分如下：

1
2
boy bo oy b o y
girl gir irl gi ir rl g i r l

输出时单个字母b o y g i r l可在和a z一同判断

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#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
using namespace std;

char str[260];

void init();
void replace_boy(char*, int);
void replace_girl(char*, int);

int main()
{
    cin >> str;
    init();
    int boy = 0, girl = 0;
    for (int i = 0, len = strlen(str); i < len; i++)
    {
        if (str[i] == 'a' || str[i] == 'b' || str[i] == 'o' || str[i] == 'y')
            boy++;
        else if (str[i] == 'z' || str[i] == 'g' || str[i] == 'i' || str[i] == 'r' || str[i] == 'l')
            girl++;
    }
    cout << boy << endl << girl;
    cin.get(); cin.get();
    return 0;
}
void init()
{
    //替换
    replace_boy("boy", 3);
    replace_boy("bo", 2);
    replace_boy("oy", 2);

    replace_girl("girl", 4);
    replace_girl("gir", 3);
    replace_girl("irl", 3);
    replace_girl("gi", 2);
    replace_girl("ir", 2);
    replace_girl("rl", 2);

}
void replace_boy(char* sub, int len)
{
    char *boy = strstr(str, sub);
    //在字符串str1中找str2，返回其首次出现的地址
    while (boy != NULL)
    {
        *boy = 'a';
        for (int i = 1; i < len; i++)
            *(boy + i) = '.';
        boy = strstr(str, sub);//找下一个
    }
}
void replace_girl(char* sub, int len)
{
    char *girl = strstr(str, sub);
    while (girl != NULL)
    {
        *girl = 'z';
        for (int i = 1; i < len; i++)
            *(girl + i) = '.';
        girl = strstr(str, sub);
    }
}