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。。

 //STATUS:C++_AC_31MS_684KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int f[][N];
 int T,n,m;

 int getfa(int a)
 {
     int i,ret=;
     for(i=;a;i++,a/=)
         ret+=(a%)*(<<i);
     return ret;
 }

 int solve()
 {
     int i,j,k,acc=,fa,len,t,ret=;
     int num[];
     for(len=,t=m;t;t/=)num[++len]=t%;
     fa=getfa(n);
     for(i=len;i>=;i--){
         for(j=;j<num[i];j++){
             if(fa-acc-j*(<<(i-))>=)ret+=f[i-][fa-acc-j*(<<(i-))];
         }
         acc+=j*(<<(i-));
     }
     if(acc<=fa)ret++;
     return ret;
 }

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,k,t,ca=;
     f[][]=;
     for(i=;i<=;i++){
         for(j=;j<;j++){
             t=j*(<<(i-));
             for(k=;t+k<N;k++)
                 f[i][t+k]+=f[i-][k];
         }
     }
     for(i=;i<=;i++)
         for(j=;j<N;j++)f[i][j]+=f[i][j-];
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         printf("Case #%d: %d\n",ca++,solve());
     }
     return ;
 }