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=10010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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[11][N];
int T,n,m;

int getfa(int a)
{
    int i,ret=0;
    for(i=0;a;i++,a/=10)
        ret+=(a%10)*(1<<i);
    return ret;
}

int solve()
{
    int i,j,k,acc=0,fa,len,t,ret=0;
    int num[13];
    for(len=0,t=m;t;t/=10)num[++len]=t%10;
    fa=getfa(n);
    for(i=len;i>=1;i--){
        for(j=0;j<num[i];j++){
            if(fa-acc-j*(1<<(i-1))>=0)ret+=f[i-1][fa-acc-j*(1<<(i-1))];
        }
        acc+=j*(1<<(i-1));
    }
    if(acc<=fa)ret++;
    return ret;
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,k,t,ca=1;
    f[0][0]=1;
    for(i=1;i<=10;i++){
        for(j=0;j<10;j++){
            t=j*(1<<(i-1));
            for(k=0;t+k<N;k++)
                f[i][t+k]+=f[i-1][k];
        }
    }
    for(i=0;i<=10;i++)
        for(j=1;j<N;j++)f[i][j]+=f[i][j-1];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        printf("Case #%d: %d
",ca++,solve());
    }
    return 0;
}