bzoj4542[HNOI2016]大数(莫队)

题目链接

BZOJ

洛谷

解析

莫队

\(num_i\)表示\(s[1…n]\)所表示的数

那么\(s[l…r]\)表示的数为\(\frac{num_l - num_{r + 1}}{10^{n - r}}\)

所以题目即求满足\(\frac{num_l - num_{r + 1}}{10^{n - r}} \equiv 0 \ (mod \ p)\)\((l, r)\)对数

  1. \(p \neq 2, p \ne 5\)时,\(p\)\(10\)互质,上式可化为\(num_l \equiv num_{r + 1} \ (mod \ p)\),问题变成了统计区间相同数个数,上莫队就行了(当然要离散化余数)
  2. \(p = 2\)\(p = 5\)时,末位数字有规律,可以继续莫队,或者前缀和搞一搞

一个注意的地方:离散化的时候把\(n+1\)位置一起离散……

代码

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define MAXN 100010

typedef long long LL;
struct Query {
	int l, r, id, blk;
	bool operator <(const Query &q) const;
};

void prework();
void work1();
void work2();
void work3();

int N, M, P, perblock, num[MAXN], vec[MAXN];
LL pow10[MAXN], f[MAXN], g[MAXN];
char s[MAXN];
Query qry[MAXN];

int main() {
	scanf("%d%s%d", &P, s + 1, &M);
	N = strlen(s + 1);
	if (P == 2) work1();
	else if (P == 5) work2();
	else work3();

	return 0;
}
bool Query::operator <(const Query &q) const { return blk == q.blk ? r < q.r : blk < q.blk; }
void work1() {
	for (int i = 1; i <= N; ++i) {
		f[i] = f[i - 1], g[i] = g[i - 1];
		if (((s[i] - '0') & 1) ^ 1) f[i] = f[i] + i, ++g[i];
	}
	while (M--) {
		int l, r; scanf("%d%d", &l, &r);
		printf("%lld\n", f[r] - f[l - 1] - (l - 1) * (g[r] - g[l - 1]));
	}
}
void work2() {
	for (int i = 1; i <= N; ++i) {
		f[i] = f[i - 1], g[i] = g[i - 1];
		if (s[i] == '0' || s[i] == '5') f[i] = f[i] + i, ++g[i];
	}
	while (M--) {
		int l, r; scanf("%d%d", &l, &r);
		printf("%lld\n", f[r] - f[l - 1] - (l - 1) * (g[r] - g[l - 1]));
	}	
}
void prework() {
	while (perblock * perblock <= N) ++perblock;
	pow10[0] = 1;
	for (int i = 1; i <= N; ++i) pow10[i] = pow10[i - 1] * 10 % P;
	for (int i = N; i; --i) num[i] = ((LL)num[i + 1] + (s[i] - '0') * pow10[N - i] % P) % P;
	for (int i = 1; i <= N; ++i) vec[i - 1] = num[i];
	std::sort(vec, vec + N + 1);//注意要把N+1位置的0一起离散化
	int tot = std::unique(vec, vec + N + 1) - vec;
	for (int i = 1; i <= N + 1; ++i)
		num[i] = std::lower_bound(vec, vec + tot, num[i]) - vec;
	for (int i = 0; i < M; ++i) {
		scanf("%d%d", &qry[i].l, &qry[i].r);
		qry[i].id = i;
		++qry[i].r;
		qry[i].blk = (qry[i].l - 1) / perblock;
	}
	std::sort(qry, qry + M);
}
void work3() {
	prework();
	LL cur = 0;
	for (int i = 0, l = 1, r = 0; i < M; ++i) {
		while (r < qry[i].r) ++r, cur += g[num[r]], ++g[num[r]];
		while (l > qry[i].l) --l, cur += g[num[l]], ++g[num[l]];
		while (r > qry[i].r) --g[num[r]], cur -= g[num[r]], --r;
		while (l < qry[i].l) --g[num[l]], cur -= g[num[l]], ++l;
		f[qry[i].id] = cur;
	}
	for (int i = 0; i < M; ++i) printf("%lld\n", f[i]);
}
//Rhein_E
posted @ 2019-03-21 18:33  Rhein_E  阅读(154)  评论(0编辑  收藏  举报