Discription
Yuuko and Nagi like to play the following game:
Initially they take a checkered sheet of paper of the size of N x M cells. Then, one cell, let‘s call this cell (X, Y) is marked. Then, they take the moves alternately. A move consists of making a long horizontal or vertical cut. This cut should coincide with one of the lines that divide the sheet into cells. Then, a part that doesn‘t contain a marked cell is thrown away and another player takes her turn. A player who is unable to make a turn because after the opponent‘s turn the dimensions of the sheet became equal to 1 x 1 loses. Yuuko plays first.
Yuuko and Nagi have played a lot of games together so now they both know the optimal strategy and always use it. Today girls are going to play a game on a N x M cells sheet of paper, but they haven‘t yet decided, what is the cell (X, Y) to be marked. Yuuko is interested, in how many ways it‘s possible to choose this cell so the she can become a winner, regardless of Nagi‘s moves.
Input
The first line of input consists of an integer T - the number of test cases. Then, Tlines describing the test cases follow. The i-th such line contains two integers Nand M, separated by a single space.
Output
For each test case, output a single line containing the number of ways to choose the marked cell in order to ensure Yuuko‘s win.
Example
Input: 2 5 8 6 7 Output: 40 42
Scoring
T = 100, 1 <= N, M <= 10 : 25 points.
T = 100, 1 <= N, M <= 1000 : 36 points.
T = 100, 1 <= N, M <= 106 : 39 points.
至于中文题面。。。
Mandarin Chinese
假设我们选了(x,y)点，那么它上面有x-1行，下面有n-x行，左边有y-1列，右边有m-y列。
而每次切割相当于让这四个量中的一个减小任意(但必须保证操作后是非负数)
这不就是Nim游戏吗？？？
但是光看出这个还不行，因为题目还要快速的算方案数。
先手必胜的情况下SG函数的Nim和是不为0的，这个不太好算，我们不妨算一下后手必胜的方案，再用总方案减去这个就是答案(总方案数显然是n*m)
然后发现其实并不是完全意义上的4堆石子，因为行的两堆和列的两堆是有相互限制的。
我们要让4堆石子的SG函数的Nim和为0，那就让行的Nim和和列的Nim和相等即可，那么我们就用a[i]记录行的Nim和==i的有多少组，
然后找列的时候累加a[i^(m-1-i)]即可。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstring>
#define ll long long
#define maxn 1000005
using namespace std;
ll tot,n,m;
int T,a[maxn];
int main(){
scanf("%d",&T);
while(T--){
scanf("%lld%lld",&n,&m);
tot=n*m;
n--,m--;
if(n>m) swap(n,m);
for(int i=0;i<=n;i++) a[i^(n-i)]++;
for(int i=0;i<=m;i++) tot-=(ll)a[i^(m-i)];
printf("%lld\n",tot);
if(T) memset(a,0,sizeof(a));
}
return 0;
}
原文地址：https://www.cnblogs.com/JYYHH/p/8361454.html