Codeforces 932 B.Recursive Queries-前缀和 (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2,...-优快云博客

本文介绍了一种递归查询算法的实现方式，通过定义两个函数f和g，并使用前缀和的方法来解决特定区间内满足条件的整数数量的问题。文章提供了一个具体的代码示例，演示了如何高效地处理大量查询请求。

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

$\text{[math]}$

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

output

input

Copy

output

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; }

转载于:https://www.cnblogs.com/ZERO-/p/9711342.html