HDU-4734 F(x) 数位DP

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

　　注意到F(x)的值比较小，所以可以先预处理所有F(x)的组合个数。f[i][j]表示 i 位数时F(x)为 j 的个数，方程容易转移:f[i][1<<i+j]=sigma( f[i][j] )。然后求一下f[i][j]的前缀和，就可以直接统计了。。

  1 //STATUS:C++_AC_31MS_684KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=10010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=100000,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 int f[11][N];
 59 int T,n,m;
 60 
 61 int getfa(int a)
 62 {
 63     int i,ret=0;
 64     for(i=0;a;i++,a/=10)
 65         ret+=(a%10)*(1<<i);
 66     return ret;
 67 }
 68 
 69 int solve()
 70 {
 71     int i,j,k,acc=0,fa,len,t,ret=0;
 72     int num[13];
 73     for(len=0,t=m;t;t/=10)num[++len]=t%10;
 74     fa=getfa(n);
 75     for(i=len;i>=1;i--){
 76         for(j=0;j<num[i];j++){
 77             if(fa-acc-j*(1<<(i-1))>=0)ret+=f[i-1][fa-acc-j*(1<<(i-1))];
 78         }
 79         acc+=j*(1<<(i-1));
 80     }
 81     if(acc<=fa)ret++;
 82     return ret;
 83 }
 84 
 85 int main(){
 86  //   freopen("in.txt","r",stdin);
 87     int i,j,k,t,ca=1;
 88     f[0][0]=1;
 89     for(i=1;i<=10;i++){
 90         for(j=0;j<10;j++){
 91             t=j*(1<<(i-1));
 92             for(k=0;t+k<N;k++)
 93                 f[i][t+k]+=f[i-1][k];
 94         }
 95     }
 96     for(i=0;i<=10;i++)
 97         for(j=1;j<N;j++)f[i][j]+=f[i][j-1];
 98     scanf("%d",&T);
 99     while(T--)
100     {
101         scanf("%d%d",&n,&m);
102         printf("Case #%d: %d
",ca++,solve());
103     }
104     return 0;
105 }