loj #2053 莫队

\(des\)

存在一个长度为 \(n\) 的数字 \(s\), 一个素数 \(P\)

\(m\) 次询问一段区间 \([l, r]\) 内的子串构成的数是 \(P\) 的倍数

\(sol\)

对于一次询问 \([l, r]\)

答案为

\[\sum_{i=l}^{r} \sum_{j=i}^{r}[(\sum_{k=i}^{j} s_{k} \times 10^{j-k}) \pmod P \equiv 0]
\]

等价于

\[\sum_{i=l}^{r} \sum_{j=i}^{r} [10^{j} (\sum_{k=i}^{j} s_{k} \times 10^{-k}) \pmod P \equiv 0]
\]

当 \(P \ne 2 且 P \ne 5\) 时，\(p \nmid 10^j\)

所以原式等价于

\[\sum_{i=l}^{r} \sum_{j=i}^{r} [(\sum_{k=i}^{j} s_{k} \times 10^{-k}) \pmod P \equiv 0]
\]

令

\(a_k = s_k \times 10^{-k} \pmod P\)

\(sum_k = \sum_{i=1}^{k} a_i \pmod P\)

所以原式等价于

\[\begin{split}
& \sum_{i=l}^{r} \sum_{j=i}^{r} [(\sum_{k=i}^{j} a_k) \pmod P \equiv 0] \\
= &\sum_{i=l}^{r} \sum_{j=i}^{r} [(sum_j = sum_{i-1})]
\end{split}
\]

对 \(sum\) 离散化后转化为区间查询相等的数的个数

莫队

对于 \(P = 2 或 P = 5\) 的情况特判即可

时间复杂度 \(O(n^{1.5} + nlogn)\)

#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + 10;

#define Rep(i, a, b) for(int i = a; i <= b; i ++)
#define LL long long

#define gc getchar()
inline int read() {
	int x = 0; char c = gc;
	while(c < '0' || c > '9') c = gc;
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
	return x;
}

int P, n, m;
char s[N];
int pos[N], block;
struct Node {
	int l, r, id;
	bool operator < (const Node a) const {
		if(pos[this-> l] == pos[a.l]) return pos[this-> r] < pos[a.r];
		return pos[this-> l] < pos[a.l];
	}
} Ask[N];

LL S[N], A[N], Sum[N], Ten[N] = {0, 10};
LL Copysum[N];
LL Tong[N], Answer[N];

LL Ksm(LL a, LL b) {
	LL ret = 1;
	while(b) {
		if(b & 1) ret = ret * a % P;
		a = a * a % P;
		b >>= 1;
	}
	return ret;
}

LL Now_ans;

inline void Cut(int x) {Tong[Sum[x]] --; Now_ans -= Tong[Sum[x]];}
inline void Add(int x) {Now_ans += Tong[Sum[x]]; Tong[Sum[x]] ++;}

void MoDui() {
	int L = Ask[1].l, R = L - 1;
	Rep(i, 1, m) {
		int l = Ask[i].l - 1, r = Ask[i].r;
		for(; L < l; L ++) Cut(L);
		for(; R > r; R --) Cut(R);
		for(; L > l; L --) Add(L - 1);
		for(; R < r; R ++) Add(R + 1);
		Answer[Ask[i].id] = Now_ans;
	}
}

LL totsum[N], totcnt[N];

void Special_Judge() {
	Rep(i, 1, n) {
		totcnt[i] = totcnt[i - 1] + ((s[i] - '0') % P == 0 ? 1 : 0);
		totsum[i] = totsum[i - 1] + ((s[i] - '0') % P == 0 ? i : 0);
	}
	Rep(i, 1, m) {
		int l =  Ask[i].l, r = Ask[i].r;
		cout << totsum[r] - totsum[l - 1] - (l - 1) * (totcnt[r] - totcnt[l - 1]) << "\n";
	}
}

int main() {
	P = read();
	scanf("%s",s + 1);
	n = strlen(s + 1);
	m = read();
	Rep(i, 1, m) Ask[i] = (Node) {read(), read(), i};
	if(P == 2 || P == 5) {
		Special_Judge(); return 0;
	}
	block = sqrt(n);
	Rep(i, 1, n) pos[i] = (i - 1) / block + 1;
	sort(Ask + 1, Ask + m + 1);
	Rep(i, 1, n) S[i] = (s[i] - '0') % P;
	Rep(i, 2, n) Ten[i] = (Ten[i - 1] * 10) % P;
	Rep(i, 1, n) A[i] = S[i] * Ksm(Ten[i], P - 2) % P;
	Rep(i, 1, n) Sum[i] = (Sum[i - 1] + A[i]) % P;
	Rep(i, 1, n) Copysum[i] = Sum[i];
	sort(Copysum + 1, Copysum + n + 1);
	Rep(i, 1, n) Sum[i] = lower_bound(Copysum, Copysum + n + 1, Sum[i]) - Copysum;
	MoDui();
	Rep(i, 1, m) cout << Answer[i] << "\n";
	return 0;
}