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[HNOI 2016]大数

时间:2018-02-21 19:01:22      阅读:176      评论:0      收藏:0      [点我收藏+]

标签:str   排序   pow   git   tor   离散化   a*   getc   span   

Description

题库链接

给你一个长度为 \(n\) ,可含前导零的大数,以及一个质数 \(p\)\(m\) 次询问,每次询问你一个大数的子区间 \([l,r]\) ,求出子区间中有多少个子串为 \(p\) 的倍数。

\(1\leq n,m\leq 100000\)

Solution

\(a_i\) 为大数第 \(i\) 位上的数值。

注意到题目是要求 \[\sum_{i=l}^r\sum_{j=i}^r\left[\sum_{k=i}^ja_k\cdot 10^{j-k}\equiv 0\pmod{p}\right]\]

我们可以将其等价变形 \[\Rightarrow\sum_{i=l}^r\sum_{j=i}^r\left[10^j\sum_{k=i}^ja_k\cdot 10^{-k}\equiv 0\pmod{p}\right]\]

由于 \(p\) 是质数,同时在 \(p\neq 2\wedge p\neq 5\) 时, \(\nexists i,10^i\equiv 0\pmod{p}\) ;且对于 \(\forall i,10^{-i}\) 存在逆元。

所以我们先讨论 \(p\neq 2\wedge p\neq 5\) 的情况。

我们不妨记 \(sum_i=\sum\limits_{j=1}^ia_j\cdot10^{-j}\) 在模 \(p\) 意义下的结果。

原式等价于 \[\begin{aligned}\Rightarrow&\sum_{i=l}^r\sum_{j=i}^r\left[\sum_{k=i}^ja_k\cdot 10^{-k}\equiv 0\pmod{p}\right]\\=&\sum_{i=l}^r\sum_{j=i}^r\left[sum_j-sum_i\equiv 0\pmod{p}\right]\\=&\sum_{i=l}^r\sum_{j=i}^r\left[sum_j=sum_i\right]\end{aligned}\]

发现这不就是莫队的板子么,求子区间内有多少数对相等。直接套板子就好了。

然后对于 \(p= 2\vee p= 5\) ,可以特判,讨论比较简单,不再赘述。

Code

注意这个代码是过不了的。但在 luogu 上强行开 O2 过了,就得过且过吧...

但是可以有这些优化:

  1. 优化莫队的计算过程,不用每次都算一遍组合数;
  2. 不用写 \(hash\) 表,直接排序离散化就好了
//It is made by Awson on 2018.2.11
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 200000;
const int MOD = 1e6+7;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(LL x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

int p, n, m, sum[N+5], inv, lim;
int tol[MOD+5];
char num[N+5];
LL ans[N+5]; 

namespace cheat2 {
    struct bittree {
        LL c[N+5];
        void add(int x, int val) {for (; x <= n; x += lowbit(x)) c[x] += val; }
        LL query(int x) {
            LL sum = 0;
            for (; x; x -= lowbit(x)) sum += c[x];
            return sum;
        }
    }T;
    int sum[N+5];
    void main() {
        scanf("%s", num+1); n = strlen(num+1);
        for (int i = 1; i <= n; i++) 
            if ((num[i]-48)%2 == 0) T.add(i, i), sum[i] = sum[i-1]+1;
            else sum[i] = sum[i-1];
        read(m); int l, r;
        while (m--) {
            read(l), read(r); writeln(T.query(r)-T.query(l-1)-1ll*(sum[r]-sum[l-1])*(l-1));
        }
    }
}
namespace cheat5 {
    struct bittree {
        LL c[N+5];
        void add(int x, int val) {for (; x <= n; x += lowbit(x)) c[x] += val; }
        LL query(int x) {
            LL sum = 0;
            for (; x; x -= lowbit(x)) sum += c[x];
            return sum;
        }
    }T;
    int sum[N+5];
    void main() {
        scanf("%s", num+1); n = strlen(num+1);
        for (int i = 1; i <= n; i++) 
            if ((num[i]-48)%5 == 0) T.add(i, i), sum[i] = sum[i-1]+1;
            else sum[i] = sum[i-1];
        read(m); int l, r;
        while (m--) {
            read(l), read(r); writeln(T.query(r)-T.query(l-1)-1ll*(sum[r]-sum[l-1])*(l-1));
        }
    }
}

struct HASH {
    int k[MOD+5];
    void clear() {memset(k, -1, sizeof(k)); }
    void insert(int x) {
        int loc = x%MOD;
        while (true) {
            if (k[loc] == -1 || k[loc] == x) {k[loc] = x; return; }
            ++loc; if (loc >= MOD) loc %= MOD;
        }
    }
    int query(int x) {
        int loc = x%MOD;
        while (true) {
            if (k[loc] == x) {return loc; }
            ++loc; if (loc >= MOD) loc %= MOD;
        }
    }
}mp;
struct tt {
    int l, r, id;
    bool operator < (const tt &b) const {return l/lim == b.l/lim ? r < b.r : l < b.l; } 
}a[N+5];
int quick_pow(int a, int b, int p) {
    int ans = 1;
    while (b) {
        if (b&1) ans = 1ll*ans*a%p;
        b >>= 1, a = 1ll*a*a%p;
    }
    return ans;
}
LL C(int n) {return 1ll*n*(n-1)/2;}
void work() {
    read(p); mp.clear();
    if (p == 2) {cheat2::main(); return; }
    if (p == 5) {cheat5::main(); return; }
    inv = quick_pow(10, p-2, p); scanf("%s", num+1); n = strlen(num+1); lim = sqrt(n);
    mp.insert(0);
    for (int i = 1, j = inv; i <= n; i++, j = 1ll*j*inv%p) {
        sum[i] = (sum[i-1]+1ll*(num[i]-48)*j%p)%p;
        mp.insert(sum[i]); 
    }
    read(m);
    for (int i = 1; i <= m; i++) {read(a[i].l), --a[i].l, read(a[i].r), a[i].id = i; }
    sort(a+1, a+m+1);
    LL now = 0; int curl = 0, curr = 0; tol[mp.query(0)] = 1;
    for (int i = 1; i <= m; i++) {
        int l = a[i].l, r = a[i].r;
        int id = mp.query(sum[curl]);
        while (curl < l) now -= C(tol[id]), --tol[id], now += C(tol[id]), ++curl, id = mp.query(sum[curl]);
        id = mp.query(sum[curl-1]);
        while (curl > l) --curl, now -= C(tol[id]), ++tol[id], now += C(tol[id]), id = mp.query(sum[curl-1]);
        id = mp.query(sum[curr+1]);
        while (curr < r) ++curr, now -= C(tol[id]), ++tol[id], now += C(tol[id]), id = mp.query(sum[curr+1]);
        id = mp.query(sum[curr]);
        while (curr > r) now -= C(tol[id]), --tol[id], now += C(tol[id]), --curr, id = mp.query(sum[curr]);
        ans[a[i].id] = now;
    }
    for (int i = 1; i <= m; i++) writeln(ans[i]);
}
int main() {
    work(); return 0;
}

[HNOI 2016]大数

标签:str   排序   pow   git   tor   离散化   a*   getc   span   

原文地址:https://www.cnblogs.com/NaVi-Awson/p/8445686.html

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