light oj 1140 - How Many Zeroes? 数位DP

时间:2023-03-09 17:28:27
light oj 1140 - How Many Zeroes?  数位DP

思路:dp[i][j]:表示第i位数,j表示是否有0.

代码如下:

 #include<iostream>
#include<stdio.h>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<cstring>
#include<vector>
#define ll long long
#define pi acos(-1.0)
#define MAX 50000
using namespace std;
int bit[];
ll dp[][],p[];
ll dfs(int pos,int m,bool f)
{
if(pos==-) return !m;
if(!f&&dp[pos][m]!=-) return dp[pos][m];
ll ans=;
int e=f?bit[pos]:;
for(int i=;i<=e;i++){
if(i==&&m){
if(!(f&&i==e)) ans+=p[pos]+dfs(pos-,,);
else{
ll t=;
for(int j=pos;j>=;j--) t=t*+bit[j];
t++;
ans+=t+dfs(pos-,,);
}
}
else ans+=dfs(pos-,m||i!=,f&&i==e);
}
if(!f) dp[pos][m]=ans;
return ans;
}
ll cal(ll n)
{
if(n<) return ;
int m=;
while(n){
bit[m++]=n%;
n/=;
}
return dfs(m-,,);
}
int main(){
int t,ca=;
ll a,b;
p[]=;
memset(dp,-,sizeof(dp));
for(int i=;i<;i++) p[i]=*p[i-];
scanf("%d",&t);
while(t--){
scanf("%lld%lld",&a,&b);
printf("Case %d: %lld\n",++ca,cal(b)-cal(a-));
}
return ;
}