时间限制 5000 ms 内存限制 65536 KB

题目描述

One day, the teacher asked Weishen to judge whether a binary string can be divided by 3 (3=(11)2) in binary system.
   "It’s too simple.", Weishen said, and solved it quickly.
   Teacher was very angry, he decided to teach Weishen some life experience.
   So he added '?' characters to the 01 string, '?' can replace either 0 or 1 respectively.
   This time he askes Weishen to calculate the expected number of substrings that can be divided by 3 if each '?' randomly replaces '0' or '1'. Note that the substrings can not contain leading zeroes, thus you should ignore substrings like (011)2. Weishen would like you to help him solve this problem.

输入格式

The input begins with a line containing a single integer T(1≤T≤50 ), indicating the number of test cases. For each test case, the first line contains one integers N(1≤N≤100000), indicating the length of the string. The next lines is the string containing Ncharacters given by the teacher. It is guaranteed that only characters '0', '1', and '?' would appear in the strings.

输出格式

For each case, please output a line containing a real number rounded to the second decimal place, indicating the answer.

Explanation of example

For the first case '01?', if this '?' represent '0', then we get '010', there are two substrings divisible by 3: '0' and '0'. If this '?' represent '1', we get '011' and there are also two substrings divizible by 3: '0' and '11'. String '011'has leading zeroes and we have to omit it. The expected number of substrings will be (2+2)/2=2.

输入样例

1
3
01?

输出样例

2.00

题意是给你一个长度最大10^5的二进制01串，其中有些位置被?模糊了，问所有的可能性里，所有的能被3整除的子串的数量的期望值是多少，期望DP。

样例解释：01？

0.5概率？是1, 011里，0和11都可以被3整除

0.5概率？是0, 010里，第一个0和第三个0都可以被3整除

所以0.5*2+0.5*2=2

因为要求被3整除，所以设dp[100000+5][3]，dp[i][j]表示以i位结尾的模3余数为j的子串的数量有多少。

然后有12种状态转移，null代表开头第一位前面没有字符，其余的都代表由前一位转移到当前位的字符变化情况

null->0,null->1,null->?

0->0,0->1,0->?

1->0,1->1,1->?

?->0,?->1,?->?

我们知道，对元二进制串A后加一位0意味着A*2，根据模运算

①A%3==1，(A*2)%3==2

②A%3==2，(A*2)%3==1

③A%3==0，(A*2)%3==0

所以状态转移方程里，转移到当前位是0的情况里，一定都都要有dp[i][1]=dp[i-1][2], dp[i][2]=dp[i-1][1]

我们又知道，对元二进制串A后加一位1意味着A*2+1，根据模运算

①A%3==1，(A*2+1)%3==0

②A%3==2，(A*2+1)%3==2

③A%3==0，(A*2+1)%3==1

转移到当前位是1的情况里，一定都要有dp[i][2]=dp[i-1][2], dp[i][0]=dp[i-1][1]

然后我们具体来看各个转移

1.我们来看0->0的情况显然，以前一个0结尾的可以被3整除的数，*2还是可以被整除，所以dp不变，比如110,1100，以第三位结尾的有110、0，以第4位结尾的，1100、0，没有任何变化。dp[i][0]=dp[i-1][0]

2.再来看1->0的情况，显然以前一个1结尾的可以被3整除的数，*2还是可以被整除，但新增加了一个单个的0，所以dp+1，dp[i][0]=dp[i-1][0]+1

3.如果是?->0的话，显然0->0和1->0各占50%，dp+0.5，dp[i][0]=dp[i-1][0]+0.5

4.如果是null->0的话，显然+1，因为增加了单个0.

我们再来看->1的情况，因为前一位%3==1和%3==2的情况的情况都讲明了，所以在此只看%3==0的部分

5.我们来看0->1的情况，*2+1，举个例子，100->1001, dp[3][0]=1, dp[4][1]=1, dp[i][1]=dp[i-1][0]

6.1->1的情况，举个例子，101->1011，dp[3][0]=0, dp[4][1]=1, dp[i][1]=dp[i-1][0]+1, 为什么都是*2+1，它却要额外dp+1呢，因为1变成11, 11可以被3整除，所以要+1

7.如果是?->1的话，显然dp+0.5，dp[i][0]=dp[i-1][0]+0.5

然后字符->?后续的情况把上面都加概率仔细算一下就好

代码略长

#include<cstdio>
#include<cmath>
#include<algorithm>
#define N 100050
using namespace std;
char str[N];
double dp[N][3];
int main(){int t,n,i;for(scanf("%d",&t);t--;){scanf("%d",&n);scanf("%s",str+1);dp[0][0]=dp[0][1]=dp[0][2]=0;str[0]='Z';double res=0;for(i=1;i<=n;i++){if(str[i]=='0'){if(str[i-1]=='0'){dp[i][0]=dp[i-1][0];}else if(str[i-1]=='1'){dp[i][0]=dp[i-1][0]+1;}else if(str[i-1]=='?'){dp[i][0]=dp[i-1][0]+0.5;}else {dp[i][0]=dp[i-1][0]+1;}dp[i][1]=dp[i-1][2];dp[i][2]=dp[i-1][1];}else if(str[i]=='1'){if(str[i-1]=='0'){dp[i][1]=dp[i-1][0];}else if(str[i-1]=='1'){dp[i][1]=dp[i-1][0]+1;//*2+1,1mod 3==1}else if(str[i-1]=='?'){dp[i][1]=dp[i-1][0]+0.5;}else{dp[i][1]=dp[i-1][0]+1;}dp[i][2]=dp[i-1][2];dp[i][0]=dp[i-1][1];//no add 0}else {dp[i][0]=0.5*dp[i-1][1];dp[i][1]=0.5*dp[i-1][2];dp[i][2]=0.5*dp[i-1][1]+0.5*dp[i-1][2];if(str[i-1]=='0'){dp[i][0]+=dp[i-1][0]*0.5;dp[i][1]+=dp[i-1][0]*0.5;}else if(str[i-1]=='1'){dp[i][0]+=(dp[i-1][0]+1)*0.5;dp[i][1]+=(dp[i-1][0]+1)*0.5;}else if(str[i-1]=='?'){dp[i][0]+=(dp[i-1][0]+0.5)*0.5;dp[i][1]+=(dp[i-1][0]+0.5)*0.5;}else{dp[i][0]+=(dp[i-1][0]+1)*0.5;dp[i][1]+=(dp[i-1][0]+1)*0.5;}}}for(i=1;i<=n;i++)res=res+dp[i][0];printf("%.2lf\n",res);}return 0;
}

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