题意
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
*/
