Google Code Jam 2014 Round 1B Problem B
二进制数位DP，涉及到数字的按位与操作。
#include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> using namespace std; #define MAX_LEN 50 long long A, B, K; int a[MAX_LEN], b[MAX_LEN], k[MAX_LEN]; long long memoize[MAX_LEN][2][2][2]; void input() { scanf("%lld%lld%lld", &A, &B, &K); } int convert(long long A, int a[]) { int i = 0; while (A) { a[i] = A & 1; A >>= 1; i++; } return i; } long long dfs(int current_bit, bool less_a, bool less_b, bool less_k) { if (current_bit == -1) { if (less_a && less_b && less_k) { return 1; } return 0; } if (memoize[current_bit][less_a][less_b][less_k] != -1) return memoize[current_bit][less_a][less_b][less_k]; bool one_a = less_a || a[current_bit] == 1; bool one_b = less_b || b[current_bit] == 1; bool one_k = less_k || k[current_bit] == 1; // a0 b0 long long ret = dfs(current_bit - 1, one_a, one_b, one_k); // a1 b0 if (one_a) { ret += dfs(current_bit - 1, less_a, one_b, one_k); } // a0 b1 if (one_b) { ret += dfs(current_bit - 1, one_a, less_b, one_k); } // a1 b1 if (one_a && one_b && one_k) { ret += dfs(current_bit - 1, less_a, less_b, less_k); } return memoize[current_bit][less_a][less_b][less_k] = ret; } int main() { int t; scanf("%d", &t); for (int i = 0; i < t; i++) { printf("Case #%d: ", i + 1); input(); memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b)); memset(k, 0, sizeof(k)); convert(A, a); convert(B, b); convert(K, k); memset(memoize, -1, sizeof(memoize)); long long ans = dfs(31, false, false, false); printf("%lld\n", ans); } return 0; }