ZOJ3798 Abs problem

Title: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3798

learned about the sequence with next_permutation violence to hit the table

You can find the pattern first.

#include <cstdio>#include<algorithm>#defineINF 0x3f3f3f3f//#defineN 100using namespacestd;intSeq[n];//Save current sequenceintMi[n], ma[n];//save minimum maximum value corresponding sequenceintCalcintN//Total Sort Number{    intp =1;  for(inti =1; I <= N; ++i) P*=i; returnp;}intMain () {intN;  while(~SCANF ("%d", &N)) { for(inti =1; I <= N; ++i) seq[i]=i; intk = Calc (n), MIN = INF, MAX =-INF;  for(inti =0; I < K; ++i) {intb; Next_permutation (Seq+1, seq + n +1);  for(inti =1; I <= N; ++i) {if(i = =1) B =Seq[i]; Elseb = ABS (b-Seq[i]); }            if(b <=MIN) {MIN=b;  for(inti =1; I <= N; ++i) mi[i]=Seq[i]; }            if(b>=MAX) {MAX=b;  for(inti =1; I <= N; ++i) ma[i]=Seq[i]; }} printf ("%d%d\n", MIN, MAX);  for(intj =1; J <= N; ++j) printf ("%d", Mi[j]); Puts ("");  for(intj =1; J <= N; ++j) printf ("%d", Ma[j]); Puts (""); }     return 0;}

Rule: The smallest in reverse order, the biggest in reverse

N%4==3 or 0 o'clock, the minimum value is 0 and the rest is 1

As an example: Sequence 7 6 5 4 3 2 1

At this point the sequence is the minimum value, it will be found that: starting from 7 per 4 number of absolute value is 0, every two number is 1, so if n%4==3, the remaining last 3 is 3 2 1, easy to get the minimum value of 0,n%4==0, 1, 2 o'clock the same.

Max value Max (n) =n-min (n-1)

AC Code:

1#include <cstdio>2 #defineN 500053 intGetmin (intN)4 {5     if(n%4==3|| N%4==0)6         return 0;7     return 1;8 }9 intMain ()Ten { One     intN; A      while(~SCANF ("%d", &N)) -     { -printf"%d%d\n", Getmin (n), N-getmin (N-1)); the          for(inti = n; i >0; --i) -         { -             if(I >1) printf ("%d", i); -             Elseprintf"%d\n", i); +         } -          for(inti = n1; i >0; --i) +printf"%d", i); Aprintf"%d\n", n); at     } -     return 0; -}

ZOJ3798 Abs problem