1.3.4 Combination Lock

Farmer John ' s cows keep escaping from his farm and causing mischief. To try and prevent them from leaving, he purchases a fancy combination lock to keep his cows from opening the pasture gate .

Knowing that he cows is quite clever, Farmer John wants to make sure they cannot easily open the lock by simply trying M Any different combinations. The lock has three dials, each numbered 1..N (1 <= N <=), where 1 and n is adjacent since the dials is Circula R. There is combinations that open the lock, one set by Farmer John, and also a "master" combination set by the lock Maker.

The lock has a small tolerance for error, however, so it would open even if the numbers on the dials is each within at MOS T 2 positions of a valid combination.

For example, if Farmer John's combination is (A-Z) and the master combination is (4,5,6), the lock would open if its dial S is set to (1,3,5) (since-is-close enough to Farmer John's combination) or to (2,4,8) (since the is close enough t o the master combination). Note that (1,5,6) would not open the lock, since it was not close enough to any one single combination.

Given Farmer John ' s combination and the master combination, please determine the number of distinct settings for the dials That'll open the lock. Order matters, so the setting (all-in-a--) is distinct from (3,2,1).

Program Name:combo

INPUT FORMAT:

Line 1:

The integer N.

Line 2:	Three space-separated integers, specifying Farmer John ' s combination.
Line 3:	Three space-separated integers, specifying the master combination (possibly the same as Farmer John ' s combination).

SAMPLE INPUT (file combo.in):

50

1 2 3

5 6 7

INPUT DETAILS:

Each dial is numbered 1..50. Farmer John's combination is (-), and the master combination is (5,6,7).

OUTPUT FORMAT:

Line 1:

The number of distinct dial settings that would open the lock.

SAMPLE OUTPUT (file combo.out):

249

SAMPLE OUTPUT Explanation

Here's a list:

1,1,1 2,2,4 3,4,2 4,4,5 5,4,8 6,5,6 7,5,9 3,50,2 50,1,4

1,1,2 2,2,5 3,4,3 4,4,6 5,4,9 6,5,7 7,6,5 3,50,3 50,1,5

1,1,3 2,3,1 3,4,4 4,4,7 5,5,5 6,5,8 7,6,6 3,50,4 50,2,1

1,1,4 2,3,2 3,4,5 4,4,8 5,5,6 6,5,9 7,6,7 3,50,5 50,2,2

1,1,5 2,3,3 3,4,6 4,4,9 5,5,7 6,6,5 7,6,8 49,1,1 50,2,3

1,2,1 2,3,4 3,4,7 4,5,5 5,5,8 6,6,6 7,6,9 49,1,2 50,2,4

1,2,2 2,3,5 3,4,8 4,5,6 5,5,9 6,6,7 7,7,5 49,1,3 50,2,5

2,4,1 3,4,9 4,5,7 5,6,5 6,6,8 7,7,6 49,1,4 50,3,1

1,2,4 2,4,2 3,5,5 4,5,8 5,6,6 6,6,9 7,7,7 49,1,5 50,3,2

1,2,5 2,4,3 3,5,6 4,5,9 5,6,7 6,7,5 7,7,8 49,2,1 50,3,3

1,3,1 2,4,4 3,5,7 4,6,5 5,6,8 6,7,6 7,7,9 49,2,2 50,3,4

1,3,2 2,4,5 3,5,8 4,6,6 5,6,9 6,7,7 7,8,5 49,2,3 50,3,5

1,3,3 3,1,1 3,5,9 4,6,7 5,7,5 6,7,8 7,8,6 49,2,4 50,4,1

1,3,4 3,1,2 3,6,5 4,6,8 5,7,6 6,7,9 7,8,7 49,2,5 50,4,2

1,3,5 3,1,3 3,6,6 4,6,9 5,7,7 6,8,5 7,8,8 49,3,1 50,4,3

1,4,1 3,1,4 3,6,7 4,7,5 5,7,8 6,8,6 7,8,9 49,3,2 50,4,4

1,4,2 3,1,5 3,6,8 4,7,6 5,7,9 6,8,7 1,50,1 49,3,3 50,4,5

1,4,3 3,2,1 3,6,9 4,7,7 5,8,5 6,8,8 1,50,2 49,3,4 49,50,1

1,4,4 3,2,2 3,7,5 4,7,8 5,8,6 6,8,9 1,50,3 49,3,5 49,50,2

1,4,5 3,2,3 3,7,6 4,7,9 5,8,7 7,4,5 1,50,4 49,4,1 49,50,3

2,1,1 3,2,4 3,7,7 4,8,5 5,8,8 7,4,6 1,50,5 49,4,2 49,50,4

2,1,2 3,2,5 3,7,8 4,8,6 5,8,9 7,4,7 2,50,1 49,4,3 49,50,5

2,1,3 3,3,1 3,7,9 4,8,7 6,4,5 7,4,8 2,50,2 49,4,4 50,50,1

2,1,4 3,3,2 3,8,5 4,8,8 6,4,6 7,4,9 2,50,3 49,4,5 50,50,2

2,1,5 3,3,3 3,8,6 4,8,9 6,4,7 7,5,5 2,50,4 50,1,1 50,50,3

2,2,1 3,3,4 3,8,7 5,4,5 6,4,8 7,5,6 2,50,5 50,1,2 50,50,4

2,2,2 3,3,5 3,8,8 5,4,6 6,4,9 7,5,7 3,50,1 50,1,3 50,50,5

2,2,3 3,4,1 3,8,9 5,4,7 6,5,5 7,5,8

Code:

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6 using namespacestd;7 BOOLnum[111][111][111];8 intT;9 intd1[3],d2[3];Ten intCheckintNintm) One {     A     if(ABS (M-N) <=2)return 1; -     if(ABS (M-N) >=t-2)return 1; -     return 0; the } - intMain () - { -Freopen ("combo.in","R", stdin); +Freopen ("Combo.out","W", stdout); -     intCnt=0; +Cin>>T; Amemset (NUM,false,sizeof(num)); at      for(intI=0;i<3; i++) cin>>D1[i]; -      for(intI=0;i<3; i++) cin>>D2[i]; -      for(intI=1; i<=t;i++){ -          for(intj=1; j<=t;j++){ -              for(intk=1; k<=t;k++){ -                 if(num[i][j][k]==false){ in                     if(Check (i,d1[0]) &&check (j,d1[1]) &&check (k,d1[2]))|| (Check (i,d2[0]) &&check (j,d2[1]) &&check (k,d2[2]))) -cnt++; tonum[i][j][k]=true; +                 } -             } the         } *     } $cout<<cnt<<Endl;Panax Notoginseng     return 0; -}

1.3.4 Combination Lock