Description
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 4 22 73 9 45 64 6 47 55 7 2 62 4 output 1 4 0 8 input 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
题意：
给定一个表达式，根据表达式求值。先输入q，代表有q次询问，接下来q行，分别为l,r,k。表示从l到r有多少个值为k的数字，输出一个整数
思路：
看数据量，q和l,r都很大，所以如果直接暴力会超时。所以要先求出每个数的值是多少。这里通过递归或者循环都能求出来，本人不习惯递归，使用循环求出。然后用二维数组，直接求出从1-i有多少个符合条件的数，二位数组的i代表从1到i有多少符合条件的，j代表的是结果，也就是k。所以求l到r之间等于k的就是ans[r][k] - ans[l-1][k]。具体看代码。
代码：
#include <bits/stdc++.h> #include <cstdlib> #include <cstring> #include <cstdio> #include <cmath> #include <iostream> #include <algorithm> #include <string> #include <queue> #include <stack> #include <map> #include <set> #define IO ios::sync_with_stdio(false);\ cin.tie(0); cout.tie(0); typedef long long LL; const long long INF = 0x3f3f3f3f; const long long mod = 1e9+7; const double PI = acos(-1.0); const int maxn = 1100000; const char week[7][10]= {"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"}; const char month[12][10]= {"Janurary","February","March","April","May","June","July", "August","September","October","November","December" }; const int daym[2][13] = {{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31} }; const int dir4[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; const int dir8[8][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {-1, -1}, {1, -1}, {-1, 1}}; using namespace std; using namespace std; int n, k, ans[maxn][9], q, l, r; int a[maxn]; void init()//将1-maxn的值求出来 { for(int i=1; i<=maxn; i++) { int sum=1; int ii=i; while(ii) { if(ii%10!=0) sum=sum*(ii%10); ii/=10; } int sum1=1; while(1) { if(sum<10) { a[i]=sum; break; } while(sum) { if(sum%10!=0) sum1=sum1*(sum%10); sum/=10; } sum=sum1; sum1=1; } } } int main() { init(); for (int i = 1; i <= maxn; i++)//将每个数的结果加上 { ans[i][a[i]]++; } for (int i = 1; i <= 9; i++) for (int j = 2; j <= maxn; j++)//将从1到j的等于i的数求出来。 ans[j][i] += ans[j-1][i]; cin >> q; while (q--) { cin >> l>> r>> k; cout <<ans[r][k] - ans[l-1][k]<<endl; } }
