Codeforces Round #289 (Div. 2, ACM ICPC Rules)E.Pretty Song
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Codeforces Round #289 (Div. 2, ACM ICPC Rules)
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[i]=s[0]+s[1]+s[2]+...+s[i]
则一式转化为sum[n]-sum[k-1]-sum[n-k]
#include<bits/stdc++.h>const int maxn=500010;typedef long long ll;using namespace std;char s[maxn];int a[maxn];ll sum[maxn];int main(){ scanf("%s",s+1); int l=strlen(s+1); for(int i=1;i<=l;++i){ a[i]=(s[i]=='I'||s[i]=='E'||s[i]=='A'||s[i]=='O'||s[i]=='U'||s[i]=='Y'); } for(int i=1;i<=l;++i){ a[i]=a[i-1]+a[i]; sum[i]=sum[i-1]+a[i]; } double ans=0.0; for(int i=1;i<=l;++i){ ans+=(sum[l]-sum[i-1]-sum[l-i])/(i*1.0); } printf("%.6lf\n",ans); return 0;}
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- Codeforces Round #289 (Div. 2, ACM ICPC Rules)E. Pretty Song
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