给定只包含0、1、?（0或1）的串s1，当i ＞= 2, s[i][j] = abs(s[i - 1][j] - s[i - 1][j + 1]), 有多少个s1，使得s[n] = “1“

题目

#include
using namespace std;
#define int long long
const int maxn = 1e6 + 5, mod = 1e9 + 7;
int a[maxn];
string s;
int qpow(int a, int b){
   int res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
     return res;
}
void solve(){
    int cnt0 = 0, cnt1 = 0;
    int n = s.size();
    bool bit = 0;
    for(int i = 0; s[i]; i++){
//         if(s[i] == '?') cnt++;
        if(((n - 1) & (i)) == i){
            if(s[i] == '?') cnt1++;
            else{
                bit ^= (s[i] - '0');
            }
        }
        else{
            if(s[i] == '?') cnt0++;
        }
    }
    if(bit == 0 && cnt1 == 0){
        cout << 0 << '\n';
        return;
    }
    int res = qpow(2, cnt0 + max(0LL, cnt1 - 1));
    cout << res << '\n';
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
//     cin >> T;
    while(cin >> s){
        solve();
    }
    return 0;
}