codeforces 476 B. Dreamoon and WiFi
http://codeforces.com/problemset/problem/476/B 题意:给出两个长度相等-且不超过10的字符串,串1只包含‘-’,'+‘。按照‘+’为1,‘-’为-1累加可以得到一个值。串2还包含若干‘?’,代表该处的值不确定,且为'+'和'-'的概率相等,都是0.5.问串2的值和串1相等的概率。 思路:我们可以扫一遍得到‘?’的个数和两个式子的差值。设问号个数为a,差值为b,那么在a个问号中需要有(a-b)/2个为‘+’(容易知道,a,b一定奇偶性相同,所以a-b一定能被2整除),根据超几何分布,概率为 c[a][(a-b)/2]*(1/2)^a; 写的时候可以先打个组合数的表。1A,开心。
/* ***********************************************
Author :111qqz
Created Time :2016年02月02日 星期二 03时32分39秒
File Name :code/cf/problem/476B.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6string s1,s2;
7int len;
8int c[20][20];
9void pre()
10{
11 ms(c,0);
12 c[1][1] = 1;
13 c[1][2] = 1;
14 c[2][1] = 1;
15 c[2][2] = 2;
16 c[2][3] = 1;
17 for ( int i =3 ; i <=15 ; i++)
18 for ( int j = 0 ; j <= i ; j++)
19 c[i][j+1] = c[i-1][j+1]+c[i-1][j];
20}
21int main()
22{
23 #ifndef ONLINE_JUDGE
24 freopen("code/in.txt","r",stdin);
25 #endif
26 pre();
1 cin>>s1>>s2;
2 len = s1.length();
3 int a=0,b=0;
4 for ( int i = 0 ; i < len ; i++)
5 {
6 if (s2[i]=='?') a++;
7 if (s1[i]=='+') b++;
8 else b--;
9 if (s2[i]=='+') b--;
10 else if (s2[i]=='-') b++;
11 }
12// cout<<"a:"<<a<<endl;
13// cout<<"b:"<<b<<endl;
14 if (a==0)
15 {
16 if (b==0) puts("1");
17 else puts("0");
18 }
19 else
20 {
21 double ans = c[a][(a-b)/2+1]*1.0;
22 for ( int i = 1 ; i <= a ; i ++) ans *=0.5;
23 printf("%.11lf\n",ans);
24 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}