样例解释 1
对于第11个样例，合法的数对如下：(0, 1)(0,1)和(1,0)(1,0)。
对于第22个样例，合法的数对如下：(0, 10)(0,10)，(2, 8)(2,8)，(3, 9)(3,9)，(8, 2)(8,2)，(9, 3)(9,3)和(10, 0)(10,0)。
#include <cstdio>
#include <cstring>
#include <algorithm>
using int64 = long long;
int64 dp[64][2][2][2][2];
int64 xs, ys, n, m, lm, rm;
// 0 <= x <= xs, 0 <= y <= ys
int64 solve(int d, int ex, int ey, int el, int er) {
if (d < 0) {
int olm = lm & 1, orm = rm & 1;
if (el && er) return olm <= 0 && 0 <= orm;
if (el) return olm <= 0;
if (er) return orm >= 0;
return 1;
}
if (dp[d][ex][ey][el][er] != -1) return dp[d][ex][ey][el][er];
int64 ret = 0;
int ox = ex ? xs >> d & 1 : 1;
int oy = ey ? ys >> d & 1 : 1;
int on = n >> d & 1;
int olm = el ? lm >> (d + 1) & 1 : 0;
int orm = er ? rm >> (d + 1) & 1 : 1;
for (int x = 0; x <= ox; ++x) {
for (int y = 0; y <= oy; ++y) {
if ((x ^ y) != on) continue;
if (x == y && olm <= 0 && 0 <= orm) {
//printf("enter d=%d x=%d y=%d m=0\n", d, x, y);
ret += solve(d - 1, ex & (x == ox), ey & (y == oy), el & (0 == olm), er & (0 == orm));
//printf("leave d=%d x=%d y=%d m=0\n", d, x, y);
}
if (x == 0 && y == 1 && olm <= 0 && 0 <= orm) {
//printf("enter d=%d x=%d y=%d m=0\n", d, x, y);
ret += solve(d - 1, ex & (x == ox), ey & (y == oy), el & (0 == olm), er & (0 == orm));
//printf("leave d=%d x=%d y=%d m=0\n", d, x, y);
}
if (x == 1 && y == 0 && olm <= 1 && 1 <= orm) {
//printf("enter d=%d x=%d y=%d m=1\n", d, x, y);
ret += solve(d - 1, ex & (x == ox), ey & (y == oy), el & (1 == olm), er & (1 == orm));
//printf("leave d=%d x=%d y=%d m=1\n", d, x, y);
}
}
}
return dp[d][ex][ey][el][er] = ret;
}
int main() {
int T;
scanf("%d", &T);
for (int cas = 1; cas <= T; ++cas) {
// 0 <= x <= xs, 0 <= y <= ys, x ^ y = n, |x - y| <= m
scanf("%lld%lld%lld%lld", &xs, &ys, &n, &m);
lm = std::max<int64>(0, n - m);
rm = n + m;
memset(dp, -1, sizeof(dp));
printf("%lld\n", solve(62, 1, 1, 1, 1));
}
return 0;
}
