Case #1: 1
Case #2: 2
Case #3: 13
简单的数位DP:先将A表示成F(x)然后统计有多少大于x就行了。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
int t,temp;
LL l,r;
int num[20];
LL dp[20][20000];
LL dfs(int pos,int sum,int flag)
{
if(pos==0)
return sum>=0;
if(sum<0) return 0;
if(!flag&&dp[pos][sum]!=-1)
return dp[pos][sum];
LL ans=0;
int ed=flag?num[pos]:9;
for(int i=0;i<=ed;i++)
{
int s=sum-i*(1<<(pos-1));
ans+=dfs(pos-1,s,flag&&i==ed);
}
if(!flag) dp[pos][sum]=ans;
return ans;
}
LL solve(LL x)
{
int pos=0;
while(x)
{
num[++pos]=x%10;
x/=10;
}
return dfs(pos,temp,1);
}
LL work(LL x)
{
LL ans=0;
int l=0;
while(x)
{
ans+=(x%10)*(1<<l);
l++;x/=10;
}
return ans;
}
int main()
{
int cas=1;CLEAR(dp,-1);
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&l,&r);
temp=work(l);
LL ans=solve(r);
printf("Case #%d: %lld\n",cas++,ans);
}
return 0;
}