uva11859 Division Game

Division game is a 2-player game. In this game, there is a matrix of
positive integers with N rows and M columns. Players make their moves
in turns. In each step, the current player selects a row. If the row
contains all 1s, the player looses. Otherwise, the player can select
any number of integers (but at least 1 and each of them should be
greater than 1) from that row and then divides each of the selected
integers with any divisor other than 1. For example, 6 can be divided
by 2, 3 and 6, but cannot be divided by 1, 4 and 5. The player who
rst makes the matrix all 1s wins. In other words, if in his/her move,
player gets the matrix with all 1s, then he/she looses. Given the
matrix, your task is to determine whether the rst player wins or not.
Assume that both of the players will play perfectly to win. Input The
rst line has a positive integer T , T  50, denoting the number of
test cases. This is followed by each test case per line. Each test
case starts with a line containing 2 integers N and M representing the
number of rows and columns respectively. Both N and M are between 1
and 50 inclusive. Each of the next N line each contains M integers.
All these integers are between 2 and 10000 inclusive. Output For each
test case, the output contains a line in the format Case # x : M ',
where x is the case number (starting from 1) and M is YES ’ when the
rst player has a winning strategy and ` NO ’ otherwise.

把一个数变成他的因数相当于拿掉若干个他的质因子，把每行的质因子总数看成一堆火柴，就相当于NIM问题。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxx=10000;int f[10010],n,m;int main(){    int T,now,ans,i,j,x,K;    for (i=2;i<=maxx;i++)    {        x=i;        for (j=2;x>1;j++)            while (x%j==0)            {                x/=j;                f[i]++;            }    }    scanf("%d",&T);    for (K=1;K<=T;K++)    {        scanf("%d%d",&n,&m);        ans=0;        while (n--)        {            now=0;            for (i=1;i<=m;i++)            {                scanf("%d",&x);                now+=f[x];            }            ans^=now;        }        printf("Case #%d: %s\n",K,ans?"YES":"NO");    }}