CF1670F Jee, You See?

题意#

给定 $n, l, r, z$，计算长度为 $n$ 的序列的数量，使得：

• $l \le \sum a_i \le r$。
• $a_1 \oplus a_2 \oplus ... \oplus a_n = z$。

对 $10 ^ 9$ 取模。

$n \le 10 ^ 3, l \le r \le 10 ^ {18}, z \le 10 ^ {18}$。

Sol#

对于 $[l, r]$ 的问题，经典数位 $\texttt{dp}$ 差分一下。

考虑数位 $\texttt{dp}$，从低位向高位确定，设 $f_{x, num, 0/1}$ 表示确定 $[x, len]$ 之间的二进制位，上一位向第 $x$ 位进了 $num$ 位，是否小于等于和的限制。

考虑枚举当前位选择的个数，乘上组合数，做完了。

复杂度：$O(n ^ 2 \times \log)$。

Code#

Copy
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#define ll long long
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
ll read() {
    ll p = 0, flg = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') flg = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        p = p * 10 + c - '0';
        c = getchar();
    }
    return p * flg;
}
void write(int x) {
    if (x < 0) {
        x = -x;
        putchar('-');
    }
    if (x > 9) {
        write(x / 10);
    }
    putchar(x % 10 + '0');
}
bool _stmer;

const int N = 1e3 + 5, M = 65, mod = 1e9 + 7;

void Mod(int &x) {
    if (x >= mod) x -= mod;
    if (x < 0) x += mod;
}

array <array <int, N>, N> C;

void init(int n) {
    C[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        C[i][0] = 1;
        for (int j = 1; j <= i; j++)
            C[i][j] = C[i - 1][j] + C[i - 1][j - 1], Mod(C[i][j]);
    }
}

array <array <array <int, 2>, N>, M> f;
array <int, M> s;

int dfs(int x, int num, int flg, int len, int n, ll res) {
    if (~f[x][num][flg]) return f[x][num][flg];
    if (x > len) return f[x][num][flg] = (!num && !flg);
    f[x][num][flg] = 0;
    for (int i = 0; i <= n; i++) {
        if (((res >> (x - 1)) & 1) != (i & 1)) continue;
        int now = ((i + num) & 1), _flg = (now < s[x]) ? 0 : (now == s[x] ? flg : 1);
        f[x][num][flg] += 1ll
                        * dfs(x + 1, (i + num) >> 1, _flg, len, n, res) * C[n][i] % mod;
        Mod(f[x][num][flg]);
    }
    /* cerr << x << " " << num << " " << flg << " " << f[x][num][flg] << endl; */
    return f[x][num][flg];
}

int solve(int n, ll x, ll k) {
    if (!x || x < k) return 0;
    for (int i = 0; i < M; i++)
        for (int j = 0; j < N; j++)
            f[i][j].fill(-1);
    int cnt = 0;
    ll tp = x;
    while (tp) {
        cnt++;
        s[cnt] = tp & 1;
        tp >>= 1;
    }
    return dfs(1, 0, 0, cnt, n, k);
}

bool _edmer;
int main() {
    cerr << (&_stmer - &_edmer) / 1024.0 / 1024.0 << "MB\n";
    init(1000);
    int n = read();
    ll l = read(), r = read(), k = read();
    /* cerr << solve(n, 0, k) << "@" << endl; */
    write((solve(n, r, k) - solve(n, l - 1, k) + mod) % mod), puts("");
    return 0;
}