BZOJ3450 Tyvj1952 Easy

貌似就是道简单的DP题。。。恩！

令f[i]表示到第i位的期望分数，l[i]表示到第i位的期望连续'o'的长度

对于长度为l的连续'o'，长度又增加1，则贡献增加(l + 1)2 - l2 = 2 * l + 1

如果第i位是'o'，则

l[i] = l[i - 1] + 1, f[i] = f[i - 1] + (2 * l[i - 1] + 1) (后面的一个等式可以推广到'?'的情况)

如果第i位是'x'，则

l[i] = 0, f[i] = f[i - 1]

如果第i位是'?'，则有一半概率是'x'，一半概率是'o'讨论一下即可

l[i] = (l[i - 1] + 1) / 2, f[i] = f[i - 1] + (2 * l[i - 1] + 1) / 2 = f[i - 1] + (l[i - 1] + 0.5)

于是就做完辣~

 

 1 /**************************************************************
 2     Problem: 3450
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:12 ms
 7     Memory:1100 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11  
12 using namespace std;
13 typedef double lf;
14 const int Maxlen = 3e5 + 100;
15  
16 int n;
17 lf f[2], l[2];
18  
19 char buf[Maxlen], *ch = buf;
20 int Len;
21  
22 inline int read() {
23   int x = 0;
24   while (*ch < '0' || '9' < *ch) ++ch;
25   while ('0' <= *ch && *ch <= '9')
26     x = x * 10 + *ch - '0', ++ch;
27   return x;
28 }
29  
30 int main() {
31   Len = fread(ch, 1, Maxlen, stdin);
32   buf[Len] = '\0';
33   int i, now;
34   n = read();
35   for (i = 1, now = 0; i <= n; ++i, ++ch) {
36     now = !now;
37     while (*ch != 'o' && *ch != 'x' && *ch != '?') ++ch;
38     if (*ch == 'o')
39       l[now] = l[!now] + 1, f[now] = f[!now] + (2 * l[!now] + 1);
40     else if (*ch == 'x')
41       l[now] = 0, f[now] = f[!now];
42     else l[now] = (l[!now] + 1) / 2.0, f[now] = f[!now] + (l[!now] + 0.5);
43   }
44   printf("%.4lf\n", f[now]);
45   return 0;
46 }
View Code

 (p.s. 突然觉得这表情好魔性'o' 'o' 'o' 'o')

posted on 2015-02-15 17:55  Xs酱~  阅读(366)  评论(0)  编辑  收藏  举报