Gym

题意：有n张标有“C”或“F”的卡片。

1、随机取前k张(1<=k<=n)

2、若这k张的第一张为“C”，则不翻转，否则，全部翻转这k张。

3、然后处理剩下的n-k张

4、重复步骤1~3直至处理完所有卡片。

求处理后卡片W的个数期望。

分析：期望dp从后往前推。

1、对于最后一张卡片，无论为C还是W，最后的结果都是W个数为0，因此dp[len] = 0.

2、对于卡片i，k有（len - i + 1）种选择

若卡片i为C，则前k张卡片是不反转的，预处理前缀和。

k = 1, 1 / (len - i + 1) * (0 + dp[i + 1])

k = 2, 1 / (len - i + 1) * (前两张卡片中W的个数 + dp[i + 2])

.......

k = len - i + 1, 1 / (len - i + 1) * (前len - i + 1张卡片中W的个数)

若卡片i为W，同理也处理处前缀和。

提取公因数1 / (len - i + 1)后，后面的式子合并dp[i + 1] + dp[i + 2] +...+ dp[len]可以边算边处理出后缀和sumsuf[i]。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char s[MAXN];
LL a1[MAXN], a2[MAXN];
LL sum1[MAXN], sum2[MAXN];
double sumsuf[MAXN], dp[MAXN];
int main(){
    freopen("foreign.in", "r", stdin);
    freopen("foreign.out", "w", stdout);
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    for(int i = 1; i <= len; ++i){
        if(s[i] == 'W') a1[i] = a1[i - 1] + 1;
        else a1[i] = a1[i - 1];
    }
    for(int i = 1; i <= len; ++i){
        if(s[i] == 'C') a2[i] = a2[i - 1] + 1;
        else a2[i] = a2[i - 1];
    }
    for(int i = 1; i <= len; ++i){
        sum1[i] = sum1[i - 1] + a1[i];
        sum2[i] = sum2[i - 1] + a2[i];
    }
    dp[len] = 0;
    sumsuf[len] = 0;
    for(int i = len - 1; i >= 1; --i){
        LL tmp;
        if(s[i] == 'C'){
            tmp = sum1[len] - sum1[i - 1] - a1[i] * (len - i + 1);
        }
        else{
            tmp = sum2[len] - sum2[i - 1] - a2[i] * (len - i + 1);
        }
        dp[i] = (1 / (double)(len - i + 1)) * (sumsuf[i + 1] + tmp);
        sumsuf[i] = sumsuf[i + 1] + dp[i];
    }
    printf("%.12lf
", dp[1]);
    return 0;
}