Strings of Power

最新推荐文章于 2016-02-29 15:42:38 发布

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最新推荐文章于 2016-02-29 15:42:38 发布

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原文链接：http://www.cnblogs.com/xinyuyuanm/p/3178027.html

本文介绍了一个字符串处理问题，即找出所有包含特定子串heavy和metal的连续子串。通过扫描输入字符串并计数这些子串出现的次数来解决此问题。提供了一个C++实现示例。

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B. Strings of Power

Volodya likes listening to heavy metal and (occasionally) reading. No wonder Volodya is especially interested in texts concerning his favourite music style.

Volodya calls a string powerful if it starts with "heavy" and ends with "metal". Finding all powerful substrings (by substring Volodya means a subsequence of consecutive characters in a string) in a given text makes our hero especially joyful. Recently he felt an enormous fit of energy while reading a certain text. So Volodya decided to count all powerful substrings in this text and brag about it all day long. Help him in this difficult task. Two substrings are considered different if they appear at the different positions in the text.

For simplicity, let us assume that Volodya's text can be represented as a single string.

Input

Input contains a single non-empty string consisting of the lowercase Latin alphabet letters. Length of this string will not be greater than 1e6 characters.

Output

Print exactly one number — the number of powerful substrings of the given string.

Sample test(s)

input

heavymetalisheavymetal

heavymetalismetal

trueheavymetalissotruewellitisalsosoheavythatyoucanalmostfeeltheweightofmetalonyou

output

Note

In the first sample the string "heavymetalisheavymetal" contains powerful substring "heavymetal" twice, also the whole string "heavymetalisheavymetal" is certainly powerful.

In the second sample the string "heavymetalismetal" contains two powerful substrings: "heavymetal" and "heavymetalismetal".

继续水，不解释，贴代码;

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
char s[2000000];
int main()
{
    while(scanf("%s",s)!=EOF)
    {
        int length=strlen(s);
        long long a=0,b=0;
        for(int i=0;i<length;i++)
        {
            if(s[i]=='h'&&s[i+1]=='e'&&s[i+2]=='a'&&s[i+3]=='v'&&s[i+4]=='y')
            {
                i+=4;
                a++;
            }
            if(s[i]=='m'&&s[i+1]=='e'&&s[i+2]=='t'&&s[i+3]=='a'&&s[i+4]=='l')
            {
                i+=4;
                b+=a;
            }
        }
        printf("%d\n",b);
    }
    return 0;
}

转载于:https://www.cnblogs.com/xinyuyuanm/p/3178027.html