[HNOI 2016]大数

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。

(1leq n,mleq 100000)

Solution

记 (a_i) 为大数第 (i) 位上的数值。

注意到题目是要求 [sum_{i=l}^rsum_{j=i}^rleft[sum_{k=i}^ja_kcdot 10^{j-k}equiv 0pmod{p} ight]]

我们可以将其等价变形 [Rightarrowsum_{i=l}^rsum_{j=i}^rleft[10^jsum_{k=i}^ja_kcdot 10^{-k}equiv 0pmod{p} ight]]

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

所以我们先讨论 (p eq 2wedge p eq 5) 的情况。

我们不妨记 (sum_i=sumlimits_{j=1}^ia_jcdot10^{-j}) 在模 (p) 意义下的结果。

原式等价于 [egin{aligned}Rightarrow&sum_{i=l}^rsum_{j=i}^rleft[sum_{k=i}^ja_kcdot 10^{-k}equiv 0pmod{p} ight]\=&sum_{i=l}^rsum_{j=i}^rleft[sum_j-sum_{i-1}equiv 0pmod{p} ight]\=&sum_{i=l}^rsum_{j=i}^rleft[sum_j=sum_{i-1} ight]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= 2vee 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('
'))
#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;
}