2020 CCPC-Wannafly Winter Camp Day2 A
题目
思路
一开始没什么思路,直到看了题解,有了一点想法,然后自己推了推就推出来了,就觉得当时的自己就是个憨批,下面是我的思路,设sum[i]是字符串1~i中出现的元音字符个数,f[i]是长度为i的子串中出现的元音字符个数,因为我们不能暴力统计每一个子串的元音字母个数(会TLE),想到统计子串【L…R】的元音个数,我们可以使用前缀和,用sum数组记录
然后就可以推出题解的式子了
而且会发现,f数组是对称的,所以
for(i = n/2+1; i <= n; i++)
{
f[i] = f[n-i+1];
ans += double(f[i]) / double(i);
}
代码
#include <iostream>
#include <cstdio>
#include <set>
#include <list>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <string>
#include <sstream>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <fstream>
#include <iomanip>
using namespace std;
#define dbg(x) cerr << #x " = " << x <<endl;
typedef pair<int, int> P;
typedef long long ll;
const int MAXN = 1e6+5;
ll sum[MAXN];
ll f[MAXN];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
string op;
cin >> op;
int n = op.size();
for(int i = 0; i < n; i++)
{
if(op[i] == 'a' || op[i] == 'e' || op[i] == 'i' || op[i] == 'o' || op[i] == 'u' || op[i] == 'y')
{
sum[i+1] = sum[i] + 1;
}
else
{
sum[i+1] = sum[i];
}
}
double ans = 0;
int i;
for(i = 1; i <= op.size()/2; i++)
{
if(i == 1)
{
f[i] = sum[op.size()];
}
else
{
f[i] = f[i-1] + sum[n-i+1] - sum[i-1];
}
ans += double(f[i]) / double(i);
}
for(; i <= n; i++)
{
f[i] = f[n-i+1];
ans += double(f[i]) / double(i);
}
ans /= ((double(n) * (n+1) / 2.0));
cout << fixed << setprecision(8) << ans << endl;
}