[数位DP]BZOJ 4513

$Solution$

从高到低DP。
设$f_{i,j,k,l}$表示考虑了前$i$位，是否等于$n$的前$i$位，是否等于$m$的前$i$位，是否等于$k$的前$i$位的数对的个数。
设$g_{i,j,k,l}$表示考虑了前$i$位，是否等于$n$的前$i$位，是否等于$m$的前$i$位，是否等于$k$的前$i$位的数对的和。
然后枚举当前位直接转移就好啦。

using namespace std;

typedef long long ll;

inline char get(void) {
  static char buf[100000], *S = buf, *T = buf;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 100000, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
template<typename T>
inline void read(T &x) {
  static char c; x = 0; int sgn = 0;
  for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
  for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
  if (sgn) x = -x;
}

int f[70][2][2][2], g[70][2][2][2];
ll n, m, K, ans;
int P, test;

template<typename T>
inline void Add(T &x, int a) {
  x += a; while (x >= P) x -= P;
  while (x < 0) x += P;
}

int main(void) {
  freopen("1.in", "r", stdin);
  freopen("1.out", "w", stdout);
  read(test);
  while (test--) {
    read(n); read(m); read(K); read(P); n--; m--;
    memset(f, 0, sizeof f);
    memset(g, 0, sizeof g);
    f[0][1][1][1] = 1; g[0][1][1][1] = 0;
    for (int i = 0; i <= 63; i++)
      for (int j = 0; j < 2; j++)
    for (int k = 0; k < 2; k++)
      for (int l = 0; l < 2; l++) {
        if (!f[i][j][k][l]) continue;
        int nn = (n >> (63 - i)) & 1,
          mm = (m >> (63 - i)) & 1,
          kk = (K >> (63 - i)) & 1;
        for (int x = 0; x < 2; x++) {
          if (j && x > nn) continue;
          for (int y = 0; y < 2; y++) {
        if (k && y > mm) continue;
        int z = x ^ y;
        if (l && z < kk) continue;
        int nj = (j && x == nn),
          nk = (k && y == mm),
          nl = (l && z == kk);
        Add(f[i + 1][nj][nk][nl], f[i][j][k][l]);
        Add(g[i + 1][nj][nk][nl], g[i][j][k][l]);
        Add(g[i + 1][nj][nk][nl], ((ll)z << (63 - i)) % P * f[i][j][k][l] % P);
          }
        }
      }
    K %= P; ans = 0;
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++)
    for (int k = 0; k < 2; k++)
      Add(ans, g[64][i][j][k] - K * f[64][i][j][k] % P);
    Add(ans, P);
    cout << ans << endl;
  }
  return 0;
}