Codeforces Round #289 (Div. 2, ACM ICPC Rules)E.Pretty Song
CodeforcesRound#289(Div.2,ACMICPCRules)求一个字符串的所有字串的权值和,每个字串的权值为元音字母的个数比上字串的长度将字串转化为01串,那么区间[l,r]的字串的权值为(s[r]-s[l-1])/(r-l+1),枚举长度k,则所有字串的权值和为Sigma(1/k*(s[k]-[s0]+s[k+1]-s[1]+...s[n]-s[n-k])) 一式令sum[