output
standard output
Let's call some positive integer classy if its decimal representation contains no more than 33 non-zero digits. For example, numbers 44, 200000200000, 1020310203 are classy and numbers 42314231, 102306102306, 72774200007277420000 are not.
You are given a segment [L;R][L;R]. Count the number of classy integers xx such that L≤x≤RL≤x≤R.
Each testcase contains several segments, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer TT (1≤T≤1041≤T≤104) — the number of segments in a testcase.
Each of the next TT lines contains two integers LiLi and RiRi (1≤Li≤Ri≤10181≤Li≤Ri≤1018).
Output
Print TT lines — the ii-th line should contain the number of classy integers on a segment [Li;Ri][Li;Ri].
Example
input
4 1 1000 1024 1024 65536 65536 999999 1000001
output
1000 1 0 2
数位DP 直接上板子
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[20];// 每一位的上限
ll dp[20][29]; //记忆化
ll dfs(ll pos,ll prev,ll limit)
{// 当前第几位 数位条件 是否受上限
ll i;
if(pos==0)
{
if(prev >= 4) return 0;
else return 1;
}
if(!limit&&dp[pos][prev] != -1) return dp[pos][prev];
ll up;
ll ans = 0;
up = limit?a[pos]:9;
for(i=0;i<=up;i++)
{
if(i==0) ans += dfs(pos-1,prev,limit && i==up);
else ans += dfs(pos-1,prev+1,limit && i==up);
}
if(!limit) dp[pos][prev] = ans;
return ans;
}
ll solve(ll x)
{
ll p = 0;
while(x)
{
a[++p] = x%10;
x /= 10;
}
return dfs(p,0,1);
}
int main()
{
ll t,a,b;
scanf("%lld",&t);
memset(dp,-1,sizeof(dp));
while(t--)
{
scanf("%lld%lld",&a,&b);
// cout<<solve(a-1)<<endl;
// cout<<solve(b)<<endl;
printf("%lld\n",solve(b) - solve(a-1));
}
return 0;
}