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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)