Divisibility by 15

http://162.105.81.212/JudgeOnline/problem?id=2969

能被15整除分成分别被3和5整除：

能被5整除末位一定是0或5，枚举两种情况，

接下来就是删数，删的个数越少，结果越大,分别有三种情况，一个不删， 删一个，删两个，依次枚举就可以

#include <iostream>using namespace std ;char str[1010] , ans[1010] ;int num[10] , len , sum ;bool found ;void Readin (){int i , digit ;scanf ("%s" , str) ;memset (num , 0 , sizeof (num)) ;len = (int)strlen (str) ;for (sum = i = 0 ; i < len ; i ++){digit = str[i] - '0' ;num[digit] ++ ;sum += digit ;}}void Combine (int tail){int tmp[10] , i , j , modular , l , n1 , n2 , one[1010] , two[1010] ;char S[1010] ;memcpy (tmp , num , sizeof (num)) ;tmp[tail] -- ;for (n1 = n2 = 0 , i = 1 ; i < 8 ; i += 3)//从小到大统计除3余1的数，为后面删数做准备while (tmp[i]){one[n1 ++] = i ;tmp[i] -- ;}for (i = 2 ; i < 9 ; i += 3)//从小到大统计除3余2的数，为后面删数做准备while (tmp[i]){two[n2 ++] = i ;tmp[i] -- ;}memcpy (tmp , num , sizeof (num)) ;tmp[tail] -- ;modular = sum % 3 ;l = len ;if (modular == 1){if (n1)tmp[one[0]] -- , l -- ;else if (n2 > 1)tmp[two[0]] -- , tmp[two[1]] -- , l -= 2 ;else//如果删一个和删两个都不能凑成答案，则无解return ;}else if (modular == 2){if (n2)tmp[two[0]] -- , l -- ;else if (n1 > 1)tmp[one[0]] -- , tmp[one[1]] -- , l -= 2 ;elsereturn ;}for (i = 0 , j = 9 ; i < l - 1 && j >= 0 ; i ++){while (j >= 0 && !tmp[j])j -- ;S[i] = j + '0' ;tmp[j] -- ;}S[i] = tail + '0' ;S[l] = '/0' ;if (!found || strcmp (S , ans) > 0)strcpy (ans , S) ;found = true ;}void Comp (){found = false ;if (num[0])//枚举末位是0和5Combine (0) ;if (num[5])Combine (5) ;}int main (){int i ;Readin () ;Comp () ;if (!found)printf ("impossible/n") ;else{for (i = 0 ; ans[i + 1] != '/0' && ans[i] == '0' ; i ++) ;//去前导0puts (ans + i) ;}return 0 ;}