数学
其实我们发现不用每个数都去试一下，只要确定每个数字有几个就可以确定这个数。所以我们先搜索一下，然后检验。
但是这样太慢了，所以我们打表。
打出1-30的结果，然后取模。
打表的程序好像弄丢了。。。就是要各种高精度
#include<bits/stdc++.h> using namespace std; typedef long long ll; int n; ll ans, p; string s[] = { "0", "0", "245025", "38118276", "0", "701565254721", "0", "13517018740330137", "1054861758265199049", "234734126209968566538", "7470486567228198747024", "3655529240602979286948564", "74706976586522786987307024", "53025203348958311874492027381", "1055707679026725117872959265049", "729803650696038636028778345751783", "17225329259435286208202532321097545", "10392367448990719529019739012358123154", "272006553269189557898804980767500371746", "130396873127602647877846824219825388617972", "4028460591965088771059406847405053201371772", "1603523894497218565486191724343889231534702375", "64657520623442242466249906292826929365340923553", "19343435636436705194168806046639915495855042504403", "769302689915124392250215975265042038457944892036471", "239684534506691040208694761530735879121194094879875543", "10791152002373774986609529818052463359685665383236441042", "2896176361911385189942286704606894921646831443702777412148", "134581880579217016257787571305079497004703413686681117431540", "34269736053164528543199304278101128211432125247436707468636178"}; int main() { scanf("%d%lld", &n, &p); for(int i = 0; i < s[n - 1].length(); ++i) ans = (ans * 10 + s[n - 1][i] - '0') % p; printf("%lld ", ans); return 0; }