题意:求0-B的满足<=F[A]的所有可能
思路:数位DP,记忆化搜索
#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> using namespace std; int A, B; int dp[20][200000]; int bit[20]; int dfs(int cur, int num, int flag) { if (cur == -1) return num >= 0; if (num < 0) return 0; if (!flag && dp[cur][num] != -1) return dp[cur][num]; int ans = 0; int end = flag?bit[cur]:9; for (int i = 0; i <= end; i++) ans += dfs(cur-1, num-i*(1<<cur), flag&&i==end); if (!flag) dp[cur][num] = ans; return ans; } int F(int x) { int tmp = 0; int len = 0; while (x) { tmp += (x%10)*(1<<len); len++; x /= 10; } return tmp; } int cal() { int len = 0; while (B) { bit[len++] = B%10; B /= 10; } return dfs(len-1, F(A), 1); } int main() { int t; int cas = 1; scanf("%d", &t); memset(dp, -1, sizeof(dp)); while (t--) { scanf("%d%d", &A, &B); printf("Case #%d: %d\n", cas++, cal()); } return 0; }
HDU - 4734 F(x) (2013成都网络赛,数位DP),布布扣,bubuko.com
HDU - 4734 F(x) (2013成都网络赛,数位DP)
原文地址:http://blog.csdn.net/u011345136/article/details/37657701