题意：从N里选n个棋子作为起手。
题解：K倍动态减法，因为a里存的是必败态，所以分成n=a[i]和n！= a[i]两个部分
https://blog.csdn.net/tbl_123/article/details/24884861
#include<bits/stdc++.h>
#include<cstring>
using namespace std;
#define LL long long
const int N = 3000000;
LL a[N],b[N];
int main()
{
int t,k;
LL n;
scanf("%d",&t);
while(t--)
{
scanf("%d%lld",&k,&n);
a[0] = b[0] = 1;
int i = 0 ,j=0;
while(a[i]<n)
{
i++;
a[i] = b[i-1]+1;
while(a[j+1]*k<a[i])
j++;
if(a[j]*k<a[i]) b[i] = b[j]+a[i];
else b[i] = a[i];
}
LL ans;
if(a[i] == n)
ans = (LL)n-i-1;
else
ans = (LL)n-i;
cout<<ans<<endl;
}
return 0;
}