ARC091.f

Takahashi and Aoki are playing a stone-taking game. Initially, there are $\mathrm{NN}$ piles of stones, and the $i$-th pile contains ${\mathrm{Ai}}^{}$stones and has an associated integer ${\mathrm{KiKi}}^{}$.

Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:

• Choose a pile. If the $i$-th pile is selected and there are $X$ stones left in the pile, remove some number of stones between $1$ and $\mathrm{floor(X/Ki)}$ (inclusive) from the pile.

The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here,$\mathrm{floor(x)}$ represents the largest integer not greater than $x$

这道题是我学会SG函数以后做的第一道题。

可以通过打表发现，SG(x)={x-floor(x/k)-1(x%k!=0),x/k(x%k==0)}

证明的话，只会感性理解的证明。

#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 300
using namespace std;
int n,i,a[N],k[N],sg[N],ans,sum;
int get(int x,int k){
    if (x%k==0) return x/k;
    if (x<k) return 0;
    if ((x-x/k-1)/k!=x/k) return get(x-x/k-1,k);
    int p=x/k,h=(x-p*k)/(p+1);
    return get(x-h*(1+(x/k)),k);
}
int main(){
    scanf("%d",&n);
    for (i=1;i<=n;i++)
        scanf("%d%d",&a[i],&k[i]),sg[i]=get(a[i],k[i]),sum=sg[i]^sum;
    if (sum==0)
        printf("Aoki
");
    else
        printf("Takahashi
");
    return 0;
}