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\n",ca++,solve());

103     }

104     return 0;

105 }