C. Classy Numbers
time limit per test
3 seconds
memory limit per test
256 megabytes
Let's call some positive integer classy if its decimal representation contains no more than 3
non-zero digits. For example, numbers 4, 200000, 10203 are classy and numbers 4231, 102306, 7277420000
are not.
You are given a segment [L;R]
. Count the number of classy integers x such that L≤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 T (1≤T≤104) — the number of segments in a testcase.Each of the next T lines contains two integers Li and Ri (1≤Li≤Ri≤1018).
Output
Print T lines — the i-th line should contain the number of classy integers on a segment [Li;Ri]
.
Example
Input
4 1 1000 1024 1024 65536 65536 999999 1000001
Output
1000 1 0 2
AC代碼：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
typedef long long ll;
int a[20];
int dp[20][2][4];
int dfs(int pos, int check, bool limit)
{
if (pos == -1) return 1;
if (!limit && dp[pos][limit][check] != -1) return dp[pos][limit][check];
if (check >= 3) return dp[pos][limit][check] = 1;
int up = limit ? a[pos] : 9;
int tmp = 0;
for (int i = 0; i <= up; i++)
{
tmp += dfs(pos - 1, check + (i ? 1 : 0), limit && i == a[pos]);
}
if (!limit) dp[pos][limit][check] = tmp;
return tmp;
}
int solve(long long x)
{
int pos = 0;
while (x)
{
a[pos++] = x % 10;
x /= 10;
}
return dfs(pos - 1, 0, true);
}
int main()
{
long long le, ri;
int T;
cin>>T;
//memset(dp,-1,sizeof dp);可優化
while (T--)
{
scanf("%lld%lld", &le, &ri);
memset(dp, -1, sizeof dp);
printf("%d\n", solve(ri) - solve(le - 1));
}
return 0;
}