Codeforces 809C [数位DP]

DescriptionDescription

这个题好像就是求

∑x1≤i≤x2,y1≤j≤y2[ai,j≤k]∗ai,j∑x1≤i≤x2,y1≤j≤y2[ai,j≤k]∗ai,j

其中

ai,j=(i−1)xor(j−1)+1ai,j=(i−1)xor(j−1)+1

那么就可以搞成两个东西

∑i∑j[i∑i∑j[i xorxor j≤k]+∑i∑j[ij≤k]+∑i∑j[i xorxor j≤k]∗(ij≤k]∗(i xorxor j)j)

好像数位DP搞一搞就好了。。。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
const int MOD = 1000000007;

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++;
}
inline void read(int &x) {
static char c; x = 0;
for (c = get(); c < '0' || c > '9'; c = get());
for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
}

int test, x1, x2, y1, y2, k, ans;
int dp1[40][2][2][2], dp2[40][2][2][2];
int A[40], B[40], C[40];

inline void add(int &x, int y) {
x += y; while (x >= MOD) x -= MOD;
}
inline void dda(int &x, int y) {
x -= y; while (x < 0) x += MOD;
}

int Calc(int X, int Y, int k) {
if (X < 0 || Y < 0 || k < 0) return 0;
memset(dp1, 0, sizeof dp1);
memset(dp2, 0, sizeof dp2);
// dp1维护第一个式子，dp2维护第二个式子
for (int i = 0; i < 31; i++) {
A[i] = X & 1; X >>= 1;
B[i] = Y & 1; Y >>= 1;
C[i] = k & 1; k >>= 1;
}
reverse(A, A + 31);
reverse(B, B + 31);
reverse(C, C + 31);
dp1[0][1][1][1] = 1;
for (int i = 0; i < 31; i++)
for (int a = 0; a < 2; a++)
for (int b = 0; b < 2; b++)
for (int c = 0; c < 2; c++)
for (int x = 0; x < 2; x++)
for (int y = 0; y < 2; y++) {
int z = x ^ y;
if (a == 1 && x == 1 && A[i] == 0) continue;
if (b == 1 && y == 1 && B[i] == 0) continue;
if (c == 1 && z == 1 && C[i] == 0) continue;// 判断DP到的是否在询问内
add(dp1[i + 1][a & (A[i] == x)][b & (B[i] == y)][c & (C[i] == z)], dp1[i][a][b][c]);
add(dp2[i + 1][a & (A[i] == x)][b & (B[i] == y)][c & (C[i] == z)], dp2[i][a][b][c]);
add(dp2[i + 1][a & (A[i] == x)][b & (B[i] == y)][c & (C[i] == z)], (ll)dp1[i][a][b][c] * (z << (30 - i)) % MOD);
}
int sum = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++) {
add(sum, dp1[31][i][j][k]);
add(sum, dp2[31][i][j][k]);
}
return sum;
}

int main(void) {
read(test);
while (test--) {
read(x1); read(y1); read(x2); read(y2); read(k);
ans = 0; x1--; y1--; x2--; y2--; k--;
add(ans, Calc(x2, y2, k));
add(ans, Calc(x1 - 1, y1 - 1, k));
dda(ans, Calc(x2, y1 - 1, k));
dda(ans, Calc(x1 - 1, y2, k));
printf("%d\n", ans);
}
return 0;
}