Codeforces 932 B.Recursive Queries-前缀和 (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined))

 

B. Recursive Queries

 

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let us define two functions f and g on positive integer numbers.

You need to process Q queries. In each query, you will be given three integers l, r and k. You need to print the number of integers xbetween l and r inclusive, such that g(x) = k.

Input

The first line of the input contains an integer Q (1 ≤ Q ≤ 2 × 105) representing the number of queries.

Q lines follow, each of which contains 3 integers l, r and k (1 ≤ l ≤ r ≤ 106, 1 ≤ k ≤ 9).

Output

For each query, print a single line containing the answer for that query.

Examples

input

Copy

4
22 73 9
45 64 6
47 55 7
2 62 4

output

1
4
0
8

input

Copy

4
82 94 6
56 67 4
28 59 9
39 74 4

output

3
1
1
5

Note

In the first example:

• g(33) = 9 as g(33) = g(3 × 3) = g(9) = 9
• g(47) = g(48) = g(60) = g(61) = 6
• There are no such integers between 47 and 55.
• g(4) = g(14) = g(22) = g(27) = g(39) = g(40) = g(41) = g(58) = 4

 

 

这道题就是直接暴力应该会超时，用前缀和处理一下就可以了(吐槽，一开始都没看懂题，本咸鱼就没读懂过带公式的题(╥╯^╰╥))

 

代码:

//B. Recursive Queries-前缀和
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stdlib.h>
using namespace std;
const int maxn=1e6+10;
int a[10][maxn];
int fun(int x){
    if(x<10)return x;
    int sum=1;
    while(x){
        sum*=x%10>0?x%10:1;
        x/=10;
    }
    return fun(sum);
}
void qianzhuihe(){
    for(int i=1;i<=1000000;i++)
       a[fun(i)][i]++;
    for(int i=1;i<10;i++){
        for(int j=1;j<=1000000;j++)
            a[i][j]+=a[i][j-1];
    }
}
int main(){
    qianzhuihe();
    int t;
    scanf("%d",&t);
    while(t--){
        int l,r,k;
        scanf("%d%d%d",&l,&r,&k);
        printf("%d\n",a[k][r]-a[k][l-1]);
    }
    return 0;
}