HDU 4734 F(x) (数位DP)

Posted 2020-08-13 dwtfukgv

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;
}