标签：
Description
Little Toojee was a happy go lucky boy. He seemed to find most, if not all, things funny. One day he read a word and started laughing a lot. Turns out that the word consisted only of the letters L and O. Whenever he saw the subsequence ‘LOL’ in the word, he laughed for 1 second. Given t strings, find out for how long Toojee laughed on seeing each string.
Input
The first line contains t queries. This is followed by t lines each containing one string S. String S consists only of capital alphabets.
Output
Output for each string on a new line.
Constraints
Sample Input
2
LOL
LOLOL
1
4
Hint
Test1: On observation, we can tell that there is only 1 occurrence of LOL.
Test2: Let the string be 0-indexed and let V = {a, b, c} denote the indices that make up the string “LOL”, where a is index of the 1st ‘L’, b is index of the ‘O’ and c is the index of the 2nd ‘L’. So, V can be {0, 1, 2}, {2, 3, 4}, {0, 1, 4} and {0, 3, 4}. We see that there are 4 occurrences of the string “LOL”.
又是求有多少种组合的题.
和上次cf那道求等比数列的非常像.
1 /************************************************************************* 2 > File Name: code/whust/#0.1/II.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年08月09日 星期日 16时00分44秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<iomanip> 10 #include<cstdio> 11 #include<algorithm> 12 #include<cmath> 13 #include<cstring> 14 #include<string> 15 #include<map> 16 #include<set> 17 #include<queue> 18 #include<vector> 19 #include<stack> 20 #define y0 abc111qqz 21 #define y1 hust111qqz 22 #define yn hez111qqz 23 #define j1 cute111qqz 24 #define tm crazy111qqz 25 #define lr dying111qqz 26 using namespace std; 27 #define REP(i, n) for (int i=0;i<int(n);++i) 28 typedef long long LL; 29 typedef unsigned long long ULL; 30 const int inf = 0x7fffffff; 31 int main() 32 { 33 int T; 34 cin>>T; 35 string st; 36 while (T--) 37 { 38 cin>>st; 39 int len = st.length(); 40 LL p=0,q=0; 41 LL ans = 0 ; 42 for ( int i = 0 ; i < len ; i ++) 43 { 44 if (st[i]==‘L‘) 45 { 46 p++; 47 ans = ans + q; 48 } 49 else 50 { 51 q = q+ p; //p为L的个数,q表示前面有"LO"组合的个数 52 } 53 } 54 cout<<ans<<endl; 55 } 56 57 return 0; 58 }
whust #0.1 I - Laughing Out Loud
标签：
原文地址：http://www.cnblogs.com/111qqz/p/4715312.html
