UVa 11859 (Nim) Division Game

Consider all the elements of M in each row as a bunch, and turn the problem into N-Heap Nim games.

Then preprocessing the number of each number of factors within 10000, and then according to the Bouton theorem in the book, calculate the N-line factor because of the number of different or and.

0 is the initiator must lose the situation, output no, otherwise output yes

1#include <cstdio>2#include <cmath>3 4 Const intMAXP =10000;5 intF[maxp +Ten];6 7 intMain ()8 {9     //freopen ("In.txt", "R", stdin);Ten  One      for(inti =2; I <= Maxp; i++)if(!F[i]) A     { -         intt =i; -          while(T <=Maxp) the         { -              for(intj = t; J <= Maxp; J + = t) f[j]++; -T *=i; -         } +     } -  +     intT; Ascanf"%d", &T); at      for(intKase =1; Kase <= T; ++Kase) -     { -         intN, M; -scanf"%d%d", &n, &m); -         intXorsum =0; -          for(inti =0; I < n; i++) in         { -             intCNT =0; to              for(intj =0; J < M; J + +) +             { -                 intx; thescanf"%d", &x); *CNT + =F[x]; $             }Panax NotoginsengXorsum ^=CNT; -         } theprintf"Case #%d:%s\n", Kase, Xorsum?"YES":"NO"); +     } A  the     return 0; +}

code June

UVa 11859 (Nim) Division Game