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
*/