P3235 [HNOI2014]江南乐

传送门

yyb巨巨的题解
没看懂为什么整除分块的时候只要做两个就够了……

//minamoto
#include<bits/stdc++.h>
#define R register int
#define fp(i,a,b) for(R i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R i=a,I=b-1;i>I;--i)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R res,f=1;char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
const int N=1e5+5;
int SG[N],vis[N];
int Sg(int x){
    if(SG[x]!=-1)return SG[x];
    for(R i=2,k;i<=x;i=k+1){
        k=x/(x/i);
        for(R j=i;j<=min(i+1,k);++j){
            R s=0;
            if((x%j)&1)s^=Sg(x/j+1);
            if((j-x%j)&1)s^=Sg(x/j);
            vis[s]=x;
        }
    }
    fp(i,0,N)if(vis[i]!=x)return SG[x]=i;
}
int main(){
//	freopen("testdata.in","r",stdin);
    int T=read(),F=read();memset(SG,-1,sizeof(SG));
    fp(i,0,F-1)SG[i]=0;
    while(T--){
        int n=read(),s=0;
        while(n--)s^=Sg(read());
        printf("%d ",(bool)s);
    }return 0;
}