CF1270F. Awesome Substrings

发表于 | 阅读数

给长度为$n$的$01字符串$，求区间长度是区间$1$个数的倍数的数量。$n\leq 2*10^5$

即求$r-l=k(pre[r]-pre[l])$个数。$pre[i]$为$[1,i]$中$1$的个数。

  • $k\leq \sqrt T,r-pre[r]=l-pre[l]$暴力记录一下。
  • $k > \sqrt T$,此时$pre[r]-pre[l]=\frac{r-l}{k}< \sqrt T$。暴力枚举每一个位置$l$,然后枚举$1$的个数。然后$O(1)$计算出可行的范围。（像我这么菜就是很多$if$)。
  • 计算可行范围的时候控制$k>\sqrt T$
代码 
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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <unordered_map>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
const int mod = 1e9 + 7;

char s[N];
int sum[N], g[N];
int cnt = 0;
int mp[N * 500];
int main()
{
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    int T = sqrt(len);
    for (int i = 1; i <= len; i++)
    {
        sum[i] = sum[i - 1] + (s[i] == '1');
        if (s[i] == '1')
            g[++cnt] = i;
    }
    ll ans = 0;
    for (int i = 1; i <= T; ++i)
    {
        for (int j = 0; j <= len; ++j)
            ans += mp[j - sum[j] * i + i * len]++;
        for (int j = 0; j <= len; ++j)
            mp[j - sum[j] * i + i * len]--;
    }

    int fi = 1;
    int num1 = len / T;
    for (int i = 1; i <= len; i++)
    {
        while (fi <= cnt && g[fi] < i)
            fi++;
        if (fi > cnt)
            continue;
        //cout << i << "----" << endl;
        for (int j = 1; j <= num1 && fi + j - 1 <= cnt; j++)
        {

            int r = (j + fi > cnt ? len + 1 : g[j + fi]);

            int q = max(g[fi + j - 1] - i + 1, (T + 1) * j);
            int p = r - (q + i - 1) - 1;
            if (p < 0)
                continue;
            //cout << j << " " << p << " " << q << endl;
            if (q % j == 0)
                ans += 1 + p / j;
            else
            {
                int se = j - q % j;
                if (p >= se)
                    ans += 1 + (p - se) / j;
            }
        }
    }
    printf("%lld\n", ans);
}