題意：For a decimal number x with n digits (A_nA_n-1A_n-2 ... A₂A₁), we define its weight as F(x) = A_n * 2^n-1 + A_n-1 * 2^n-2 + ... + A₂ * 2 + A₁ * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).（懶得翻譯了哈哈哈）

題解：由於這裏要比較的是每個數位的帶權和，我們定義dp[i][j]為從i開始枚舉，和為j的時候有多少滿足題目意思的解

代碼：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
using namespace std;
int a[20];
int dp[15][5100];
int len;
int getsum(int x)
{
    int tsum=0;
    int f=1;
    while(x)
    {
        tsum+=(x%10)*f;
        f*=2;
        x/=10;
    }
    return tsum;
}
 

int dfs(int pos,int sta,int key,bool limit)
{
    if(pos==-1) return 1;
    if(!limit && dp[pos][sta]!=-1 ) return dp[pos][sta];
    int up=limit ? a[pos] : 9;
    int sum=0;
   // cout<<pos<<" "<<up<<endl;

    for(int i=0;i<=up;i++)
    {
        int temp=sta+i*pow(
 
2,pos);
       // cout<<temp<<endl;
        if(temp <= key)
        {
            sum+=dfs(pos-1,temp,key,limit & i==a[pos]);
        }
    }
    if(!limit) dp[pos][sta]=sum;
  //  cout<<"pos: "<<pos<<" "<<"sta :"<<sta<<" "<<"sum :"<<sum<<endl;
    return sum;
}
int solve(int tempa,int b)
{
    int akey=getsum(tempa);
  //  cout<<akey<<endl;
    int pos=0;
    while(b)
    {
        a[pos++]=b%10;
        b/=10;
    }
    len=pos-1;
    return dfs(len,0,akey,true);
}

int main()
{
   int t,Case=0;
   cin>>t;
   while(t--)
   {
       memset(dp,-1,sizeof(dp));
       int aa,b;
       cin>>aa>>b;
       printf("Case #%d: %d\n",++Case,solve(aa,b));
   }
    return 0;
}

但是，這個會TLE,沒錯，會TLE.....

仔細看一下，每次我們都要初始話dp數組，代價有點大，但是按我們剛才的思路來的話，必須每次求的時候都初始話一次

換個思路，dp[i][j]表示枚舉到i這個位置的時候，還需要j才能滿足條件，這個時候dp數組就和輸入的數無關了（em..建議多理理這個不會太好理解）

AC代碼:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
using namespace std;
int a[20];
int dp[15][5100];
int len;
int getsum(int x)
{
    int tsum=0;
    int f=1;
    while(x)
    {
        tsum+=(x%10)*f;
        f*=2;
        x/=10;
    }
    return tsum;
}
int dfs(int pos,int sta,bool limit)// 還需要sta
{
    if(pos==-1) return sta>=0;
    if(sta < 0) return 0;
    if(!limit && dp[pos][sta]!=-1 ) return dp[pos][sta];
    int up=limit ? a[pos] : 9;
    int sum=0;
    for(int i=0;i<=up;i++)
    {
        int temp=sta-i*(1<<pos);
       // cout<<temp<<endl;
        sum+=dfs(pos-1,temp,limit & i==a[pos]);
    }
    if(!limit) dp[pos][sta]=sum;
  //  cout<<"pos: "<<pos<<" "<<"sta :"<<sta<<" "<<"sum :"<<sum<<endl;
    return sum;
}
int solve(int tempa,int b)
{
    int akey=getsum(tempa);
    int pos=0;
    while(b)
    {
        a[pos++]=b%10;
        b/=10;
    }
    len=pos-1;
    return dfs(len,akey,true);
}

int main()
{
   int t,Case=0;
   scanf("%d",&t);
   memset(dp,-1,sizeof(dp));
   while(t--)
   {
      //
       int aa,b;
       scanf("%d %d",&aa,&b);
       printf("Case #%d: %d\n",++Case,solve(aa,b));
   }
    return 0;
}

hdu 4734 數位dp