Petition Game Learning

Nim game n heap matches, each pile has a number of matches, two people take turns, each can choose a pile at least one, can also take the whole heap away, unable to take the person lose.

The number of matches per pile is different or ==0, the initiator loses, otherwise the initiator wins.

http://acm.hust.edu.cn/vjudge/problem/32746 UVA11859

Test instructions: To a 2-D matrix, each time you can select 1 or more integers greater than 1 in a row of a matrix, each of them becomes a true factor. 12 can be changed to 1 2 3 4 6.

Solution: Equivalent to take off one or more elements, a row corresponding to a bunch of matches, each element of each factor as a match, that is, Nim game.

1#include <bits/stdc++.h>2 #defineMT (A, b) memset (A,b,sizeof (a))3 using namespacestd;4 Const intm=1e4+Ten;5 inta[ -][ -];6 Charanswer[2][8]={"NO","YES"};7vector<int>Prime;8 BOOL  is[M];9 intn,m;Ten voidinit () { OneMt is,0); A      for(intI=2; i*i<m;i++){ -         if( is[i])Continue; -          for(intj=i*i;j<m;j+=i) { the              is[j]=true; -         } -     } - prime.clear (); +      for(intI=2; i<m;i++){ -         if( is[i])Continue; + Prime.push_back (i); A     } at //printf ("%d", Prime[prime.size ()-1]); - } - intsolve () { -     intnim=0; -      for(intI=0; i<n;i++){ -         intsum=0; in          for(intj=0; j<m;j++){ -              for(intk=0; K<prime.size (); k++){ to                 if(Prime[k]>a[i][j])Continue; +                  while(a[i][j]%prime[k]==0){ -sum++; theA[i][j]/=Prime[k]; *                 } $             }Panax Notoginseng         } -nim^=sum; the     } +     returnnim!=0; A } the intMain () { + init (); -     intT; $      while(~SCANF ("%d",&t)) { $         intcas=1; -          while(t--){ -scanf"%d%d",&n,&m); the              for(intI=0; i<n;i++){ -                  for(intj=0; j<m;j++){Wuyiscanf"%d",&a[i][j]); the                 } -             } Wuprintf"Case #%d:%s\n", cas++, Answer[solve ()]); -         } About     } $     return 0; -}

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Petition Game Learning