Codeforces Round #625 (Div. 1, based on Technocup 2020 Final Round) D. Reachable Strings

可以发现一个 \(0\) 移动之后奇偶性不变
而如果出现连续 \(0\) 则它们俩的奇偶性不能发生改变
那么本质就是查询区间内有多少个 \(0\) 以及它们的奇偶性
根据奇偶性进行哈希即可

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) >> 1)
#define lowbit(i) ((i) & (-i))
#define ll long long
#define ull unsigned long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 998244353;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 2e5 + 7;
std::pii operator + (const std::pii &a, const std::pii &b) {
	return std::pii((a.fi + b.fi) % MOD, (a.se + b.se) % MOD);
}
std::pii operator - (const std::pii &a, const std::pii &b) {
	return std::pii((a.fi - b.fi + MOD) % MOD, (a.se - b.se + MOD) % MOD);
}
std::pii operator * (const std::pii &a, const std::pii &b) {
	return std::pii(1LL * a.fi * b.fi % MOD, 1LL * a.se * b.se % MOD);
}
std::pii operator + (const std::pii &a, int b) {
	return std::pii((a.fi + b) % MOD, (a.se + b) % MOD);
}

int n, cnt[N];
char s[N];
std::pii ha[N][2], base[N];
const std::pii Base = std::pii(233, 2333);
std::pii get(int l, int r, int opt) {
	return ha[r][opt] - ha[l - 1][opt] * base[cnt[r] - cnt[l - 1]];
}

int main() {
#ifdef LOCAL
	freopen("ans.out", "w", stdout);
#endif
	scanf("%d%s", &n, s + 1);
	base[0] = std::pii(1, 1);
	rep (i, 1, n + 1) {
		if (s[i] == '0') {
			cnt[i] = cnt[i - 1] + 1;
			ha[i][0] = ha[i - 1][0] * Base + ('0' + (i & 1));
			ha[i][1] = ha[i - 1][1] * Base + ('0' + ((i & 1) ^ 1));
		} else {
			cnt[i] = cnt[i - 1];
			ha[i][0] = ha[i - 1][0];
			ha[i][1] = ha[i - 1][1];
		}
		base[i] = base[i - 1] * Base;
	}
	int q;
	scanf("%d", &q);
	rep (__, 0, q) {
		int x1, x2, l;
		scanf("%d%d%d", &x1, &x2, &l);
		puts(get(x1, x1 + l - 1, x1 & 1) == get(x2, x2 + l - 1, x2 & 1) ? "Yes" : "No");
	}
	return 0;
}