CF1930D1 - Sum over all Substrings (Easy Version)

對於每一個 \(f(i, j)\)，我們考慮如何計算。我們發現，\(\texttt{1010}\) 式的字串很有用，所以這啟發我們如果遇到了一個模式 \(p_i = \texttt{'1'}\)，那麼我們可以在 \(i + 1\) 的位置放一個 \(\texttt{'1'}\)。這樣我們直接處理了 \(i, i + 1, i + 2\)。容易證明這是最優的。

#include <bits/stdc++.h>
using namespace std;

const int N = 3e5 + 7;

int read() {
    char c = getchar();
    int x = 0, p = 1;
    while ((c < '0' || c > '9') && c != '-') c = getchar();
    if (c == '-') p = -1, c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x * p;
}

int n;
string s;

int f(string p) {
	int res = 0;
	for (int i = 0; i < (int) p.size(); i ++) {
		if (p[i] == '1') {
			res ++;
			i += 2;
		}
	}
	return res;
}

void solve() {
	n = read();
	cin >> s; long long ans = 0;
	for (int i = 0; i < n; i ++) {
		string t = "";
		for (int j = i; j < n; j ++) {
			t += s[j];
			ans += f(t);
		}
	}
	cout << ans <<'\n';
}

signed main() {
	int t = read();
	while (t --) solve();
	return 0;
}