暑假培训选拔赛的一道题
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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
trueheavymetalissotruewellitisalsosoheavythatyoucanalmostfeeltheweightofmetalony
ou
output
3
2
3
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".
结题报告:
简而言之,本题就是对在一个字符串中每出现一个字串mental可以和前面出现的heavy匹配的个数求和
由于过程简单,直接贴代码
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
char a[1000000];
__int64 i,k,b,sum;
while(cin>>a)
{
k=strlen(a);
b=0;sum=0;
for(i=0;i<k;i++)
{
if(a[i]=='h'&&a[i+1]=='e'&&a[i+2]=='a'&&a[i+3]=='v'&&a[i+4]=='y')
{i+=4;b++;}
else {
if(a[i]=='m'&&a[i+1]=='e'&&a[i+2]=='t'&&a[i+3]=='a'&&a[i+4]=='l')
{sum=sum+b;i+=4;}
}
}
printf("%I64d\n",sum);
}
return 0;
}
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