cf 1217c

https://codeforc.es/problemset/problem/1217/C

 

这题明显是$O(nlogn)$的做法，因为二进制的位置还是很有限的，首位连续的0我们可以用前缀和来判断。 

#include <bits/stdc++.h>
using namespace std;
int const N = 200000 + 10;
int sum[N], t;
char s[N];
int main()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%s", s + 1);
        int len = strlen(s + 1);
        for (int i = 1; i <= len; i++)
            sum[i] = sum[i - 1] + (s[i] == '1');
        int ans = 0;
        for (int i = 1; i <= len; i++)
        {
            int k = 0;
            for (int j = i; j >= 1; j--)
            {
                int p = i - j;
                if ((1 << p) + k > i)
                {
                    if (k > p && k <= i && sum[j] - sum[i - k] == 0)
                        ans++;
                    break;
                }
                if (s[j] == '1')
                    k += (1 << p);
                if (p + 1 == k)
                    ans++;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}