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HDU 4734 F(x) (数位DP)
题意:给定 F(x)的不表达,给定一个 n 问 1- n中有多少数是小于等于 F(m)的。
析:dp[i][j] 表示前 i 位不大于 j 个的数量。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e5 + 5;const LL mod = 10000000000007;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}int a[15], f[15];int dp[15][11000];int dfs(int pos, int val, bool ok){ if(!pos) return 1; int &ans = dp[pos][val]; if(!ok && ans >= 0) return ans; int res = 0, n = ok ? a[pos] : 9; for(int i = 0; i <= n; ++i) if(val >= i * f[pos]) res += dfs(pos-1, val - i*f[pos], ok && i == n); if(!ok) ans = res; return res;}int solve(){ int len = 0; int k = 0; while(m){ k += (m % 10) << len; ++len; m /= 10; } len = 0; while(n){ a[++len] = n % 10; n /= 10; } return dfs(len, k, true);}int main(){ memset(dp, -1, sizeof dp); f[1] = 1; for(int i = 2; i < 12; ++i) f[i] = f[i-1] * 2; int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ scanf("%d %d", &m, &n); printf("Case #%d: %d\n", kase, solve()); } return 0;}
HDU 4734 F(x) (数位DP)
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