BZOJ4513: [Sdoi2016]储能表(数位dp)

题意

题目链接

sol

一点思路都没有，只会暴力，没想到标算是数位dp？？orz

首先答案可以分成两部分来统计

设

\[ f_{i,j}= \begin{aligned} i\oplus j &\left( i\oplus j >k\right) \\ 0 &\left( i\oplus j <=k\right) \end{aligned} \]

那么我们要求的就是

\[\sum_{i=0}^{n - 1} \sum_{j = 0}^{m - 1} f(i, j) - k * \sum_{i = 0}^{n - 1} \sum_{j = 0}^{m - 1} [f(i, j)]\]

也就是说，我们要统计出满足条件的数的异或和以及满足条件的数的对数

考虑直接在二进制下数位dp，注意这里我们要记三维状态

\(f[len][0/1][0/1][0/1]\)表示此时到第\(len\)位，是否顶着\(n\)的上界，是否顶着\(m\)的上界，是否顶着\(k\)的下界

然后直接dp就可以了

// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define pair pair<ll, ll>
#define mp make_pair 
#define fi first
#define se second 
#define ll long long 
#define int long long 
using namespace std;
const int maxn = 233;
inline ll read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
ll n, m, k, mod, lim, vis[maxn][2][2][2];
pair f[maxn][2][2][2];
void add2(ll &x, ll y) {
    if(x + y < 0) x = (x + y + mod);
    else x = (x + y >= mod ? x + y - mod : x + y);
}
ll add(ll x, ll y) {
    if(x + y < 0) return x + y + mod;
    return x + y >= mod ? x + y - mod : x + y;
}
ll mul(ll x, ll y) {
    return 1ll * x % mod * y % mod;
}
int get(ll x) {
    int len = 0; while(x) x >>= 1, len++; return len;
}
pair dfs(int now, int f1, int f2, int f3) {
    if(now > lim) return mp(0, 1);
    if(vis[now][f1][f2][f3]) return f[now][f1][f2][f3];
    vis[now][f1][f2][f3] = 1;
    pair ans = mp(0, 0);
    int l1 = (n >> lim - now) & 1, l2 = (m >> lim - now) & 1, l3 = (k >> lim - now) & 1;
    //cout << (f1 &&(!l1)) << endl;
    for(int i = 0; i <= (f1 ? l1 : 1); i++) {
        for(int j = 0; j <= (f2 ? l2 : 1); j++) {
            if(f3 && ((i ^ j) < l3)) continue;
            pair nxt = dfs(now + 1, f1 && (i == l1), f2 && (j == l2), f3 && ((i ^ j) == l3));
            add2(ans.se, nxt.se);
            add2(ans.fi, add(nxt.fi, mul(nxt.se, mul((i ^ j), (1ll << lim - now)))));
        }
    }
    return f[now][f1][f2][f3] = ans;
} 
int solve() {
    memset(vis, 0, sizeof(vis));
    memset(f, 0, sizeof(f));
    lim = 0;
    n = read(); m = read(); k = read(); mod = read(); n--; m--;
    lim = max(get(n), max(get(k), get(m)));
    pair ans = dfs(1, 1, 1, 1);
    return add(ans.fi, -mul(k, ans.se));
}
signed main() {
    for(int t = read(); t; t--, printf("%lld\n", solve()));
    return 0;
}
/*
5000
504363800392059286 554192717354508770 21453916680846604 401134357
*/