HDU 5375 Gray code

题意：给出一个二进制数，其中有些位的数字不确定，对于所有对应的格雷码，与一个序列a对应，第i位数字为1时得分a[i]，求最大的得分。

解法：一个二进制数x对应的格雷码为x ^ (x >> 1)，题解说是个dp……但其实就四种情况……判一下就好了

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;
string s;
LL a[200005];
int main()
{
    int T, cse;
    while(~scanf("%d", &T))
    {
        cse = 1;
        while(T--)
        {
            cin >> s;
            int n = s.size();
            string s1 = '0' + s;
            s1.erase(s1.end() - 1);
            for(int i = 0; i < n; i++)
                scanf("%lld", &a[i]);
            char st = '0';
            LL ans = 0;
            for(int i = 0; i < n; i++)
            {
                if(s[i] != '?')
                {
                    if(s[i] != s1[i])
                        ans += a[i];
                    st = s[i];
                }
                else
                {
                    int cnt = 0;
                    LL sum = 0, minn = a[i];
                    while(i < n && s[i] == '?')
                    {
                        cnt++;
                        sum += a[i];
                        minn = min(minn, a[i]);
                        i++;
                    }
                    char ed;
                    if(i < n)
                    {
                        ed = s[i];
                        sum += a[i];
                        minn = min(minn, a[i]);
                    }
                    else
                    {
                        ans += sum;
                        break;
                    }
                    if(st == ed)
                    {
                        if(cnt & 1)
                            ans += sum;
                        else
                            ans += sum - minn;
                    }
                    else
                    {
                        if(cnt & 1)
                            ans += sum - minn;
                        else
                            ans += sum;
                    }
                    st = s[i];
                }
            }
            printf("Case #%d: %lld
", cse++, ans);
        }
    }
    return 0;
}