[HDOJ4734]F(x)（数位dp）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4734

题意：给定了一个函数，求[0,B]内的数的个数，使得他们换算出来的结果小于F(A)。

数位dp，dp(l,s)表示l位数的值，s为被F(A)减后还剩下的值。如果s<0说明此时的F(x)比F(A)大了，要剪枝掉。直到位数被枚举完全，判断s是否大于等于0。由于这个s最大为9*(∑i=0,9 2^i)≈5120，所以s这一维可以开5120。

#include <bits/stdc++.h>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 10;
const int maxm = 5120;
int digit[maxn];
int dp[maxn][maxm];

int f(int x) {
  int ret = 0, cnt = 0;
  while(x) {
    ret += (x % 10) * (1<<cnt++);
    x /= 10;
  }
  return ret;
}

int dfs(int l, int s, bool flag) {
  if(l == 0) return s >= 0;
  if(s < 0) return 0;
  if(!flag && ~dp[l][s]) return dp[l][s];
  int ret = 0;
  int pos = flag ? digit[l] : 9;
  Rep(i, pos+1) {
    ret += dfs(l-1, s-i*(1<<(l-1)), flag&&(i==pos));
  }
  if(!flag) dp[l][s] = ret;
  return ret;
}

int solve(int x, int y) {
  int pos = 0;
  while(x) {
    digit[++pos] = x % 10;
    x /=10;
  }
  return dfs(pos, y, true);
}

int a, b;

signed main() {
  //FRead();
  int T, _ = 1;
  Rint(T);
  Clr(dp, -1);
  W(T) {
    Rint(a); Rint(b);
    a = f(a);
    printf("Case #%d: %d
", _++, solve(b, a));
  }
  RT 0;
}