[HDOJ4734]F(x)（数位dp）

Posted 2020-08-09

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。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define fr first
 4 #define sc second
 5 #define cl clear
 6 #define BUG puts("here!!!")
 7 #define W(a) while(a--)
 8 #define pb(a) push_back(a)
 9 #define Rint(a) scanf("%d", &a)
10 #define Rll(a) scanf("%I64d", &a)
11 #define Rs(a) scanf("%s", a)
12 #define Cin(a) cin >> a
13 #define FRead() freopen("in", "r", stdin)
14 #define FWrite() freopen("out", "w", stdout)
15 #define Rep(i, len) for(int i = 0; i < (len); i++)
16 #define For(i, a, len) for(int i = (a); i < (len); i++)
17 #define Cls(a) memset((a), 0, sizeof(a))
18 #define Clr(a, x) memset((a), (x), sizeof(a))
19 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
20 #define lrt rt << 1
21 #define rrt rt << 1 | 1
22 #define pi 3.14159265359
23 #define RT return
24 #define lowbit(x) x & (-x)
25 #define onecnt(x) __builtin_popcount(x)
26 typedef long long LL;
27 typedef long double LD;
28 typedef unsigned long long ULL;
29 typedef pair<int, int> pii;
30 typedef pair<string, int> psi;
31 typedef pair<LL, LL> pll;
32 typedef map<string, int> msi;
33 typedef vector<int> vi;
34 typedef vector<LL> vl;
35 typedef vector<vl> vvl;
36 typedef vector<bool> vb;
37 
38 const int maxn = 10;
39 const int maxm = 5120;
40 int digit[maxn];
41 int dp[maxn][maxm];
42 
43 int f(int x) {
44   int ret = 0, cnt = 0;
45   while(x) {
46     ret += (x % 10) * (1<<cnt++);
47     x /= 10;
48   }
49   return ret;
50 }
51 
52 int dfs(int l, int s, bool flag) {
53   if(l == 0) return s >= 0;
54   if(s < 0) return 0;
55   if(!flag && ~dp[l][s]) return dp[l][s];
56   int ret = 0;
57   int pos = flag ? digit[l] : 9;
58   Rep(i, pos+1) {
59     ret += dfs(l-1, s-i*(1<<(l-1)), flag&&(i==pos));
60   }
61   if(!flag) dp[l][s] = ret;
62   return ret;
63 }
64 
65 int solve(int x, int y) {
66   int pos = 0;
67   while(x) {
68     digit[++pos] = x % 10;
69     x /=10;
70   }
71   return dfs(pos, y, true);
72 }
73 
74 int a, b;
75 
76 signed main() {
77   //FRead();
78   int T, _ = 1;
79   Rint(T);
80   Clr(dp, -1);
81   W(T) {
82     Rint(a); Rint(b);
83     a = f(a);
84     printf("Case #%d: %d\n", _++, solve(b, a));
85   }
86   RT 0;
87 }