这个题十分简单,关键是F(x)定义的那种感觉
#include <iostream>
#include <cstring>
#include <cstdio>
#include <bits/stdc++.h>
using namespace std;
int b[21];
int l,r;
int dp[21][50001];
int F(int x)
{
int tot=1;
int ans=0;
while(x)
{
ans+=tot*(x%10);
x/=10;
tot<<=1;
}
return ans;
}
int dfs(int pos,int preok,int sup)
{
if(pos==-1) return 1;
if(preok&&dp[pos][sup]!=-1) return dp[pos][sup];
int up=preok?9:b[pos],sum=0;
for(int i=0;i<=up;i++)
{
if(sup-i*(1<<pos)>=0)
sum+=dfs(pos-1,preok||i<b[pos],sup-i*(1<<pos));
}
if(preok) return dp[pos][sup]=sum;
return sum;
}
int solve(int R)
{
int tot=0;
while(R)
{
b[tot++]=R%10;
R/=10;
}
int sum=dfs(tot-1,0,F(l));
return sum;
}
int main()
{
int n;
memset(dp,-1,sizeof(dp));
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>l>>r;
printf("Case #%d: %d\n",i,solve(r));
}
}