Manthan, Codefest 17: E. Salazar Slytherin's Locket（数位DP）

E. Salazar Slytherin's Locket

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Harry came to know from Dumbledore that Salazar Slytherin's locket is a horcrux. This locket was present earlier at 12 Grimmauld Place, the home of Sirius Black's mother. It was stolen from there and is now present in the Ministry of Magic in the office of Dolorous Umbridge, Harry's former Defense Against the Dark Arts teacher.

Harry, Ron and Hermione are infiltrating the Ministry. Upon reaching Umbridge's office, they observed a code lock with a puzzle asking them to calculate count of magic numbers between two integers l and r (both inclusive).

Harry remembered from his detention time with Umbridge that she defined a magic number as a number which when converted to a given base b, all the digits from 0 to b - 1 appear even number of times in its representation without any leading zeros.

You have to answer q queries to unlock the office. Each query has three integers bi, li and ri, the base and the range for which you have to find the count of magic numbers.

Input

First line of input contains q (1 ≤ q ≤ 105) — number of queries.

Each of the next q lines contain three space separated integers bi, li, ri (2 ≤ bi ≤ 10, 1 ≤ li ≤ ri ≤ 1018).

Output

You have to output q lines, each containing a single integer, the answer to the corresponding query.

Examples

input

22 4 93 1 10

output

12

input

22 1 1005 1 100

output

214

题意：

给出三个数b, l, r，求出从l到r中有多少数是魔法数

魔法数的定义是：b进制下所有数字出现次数都是偶数次

数位DP，定义dp[b][len][c0][c1][c2][c3][c4][c5][c6][c7][c8][c9]

表示b进制下当前位是len，在第i个数字已经出现ci次情况下有多少个数是魔法数（其中奇数次ci=1，偶数次ci=0）

PS：其实高维寻址比较慢

#include<stdio.h>#include<string.h>#define LL long longLL str[100], c[11], dp[11][66][2][2][2][2][2][2][2][2][2][2];LL Sech(LL b, LL len, LL flag, LL p){LL i, d, ans;if(flag==0 && p==0 && dp[b][len][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]]!=-1)return dp[b][len][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]];if(flag==0)d = b-1;elsed = str[len];if(len==0){for(i=0;i<=b;i++){if(c[i])break;}if(i==b+1)dp[b][len][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]] = 1;elsedp[b][len][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]] = 0;return dp[b][len][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]];}ans = 0;for(i=0;i<=d;i++){if(p==1 && i==0){if(len!=1)ans += Sech(b, len-1, 0, 1);}else{c[i] ^= 1;ans += Sech(b, len-1, i==d && flag, 0);c[i] ^= 1;}}if(flag==0 && p==0)dp[b][len][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]] = ans;return ans;}LL Jud(LL b, LL x){LL len;if(x==0)return 0;len = 0;memset(c, 0, sizeof(c));while(x){str[++len] = x%b;x /= b;}return Sech(b, len, 1, 1);}int main(void){LL T, b, L, R;scanf("%lld", &T);memset(dp, -1, sizeof(dp));while(T--){scanf("%lld%lld%lld", &b, &L, &R);printf("%lld\n", Jud(b, R)-Jud(b, L-1));}return 0;}/*110 11 11*/