传送门
神奇的bitset()
我太弱才学会bitset 。用来制造一个超大的二进制数存东西，对于这道题就是用来存那个数能被求出。bitset还用来加速求的过程，bitset|=bitset<<x[i]就直接把原来的数加上新数了（就这里最神奇）
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<bitset>
#define mod 10007
#define ll long long
#define Mod 1000007
using namespace std;
ll t,n,m,a,b,c,y,d,e,f,s[200005];
bitset<1000010> ans;
inline ll read()
{ll sum=0,f=1;char x=getchar();while(x<'0'||x>'9'){if(x=='-')f=-1;x=getchar();}while(x>='0'&&x<='9')sum=sum*10+x-'0',x=getchar();return sum*f;
}
int yjn()
{freopen("animalcupid.in","r",stdin);freopen("animalcupid.out","w",stdout);t=read();while(t--){ans.reset();ans.set(0);n=read();m=read();s[0]=read()%mod;a=read()%mod;b=read()%mod;c=read()%mod;y=read()%Mod;d=read()%Mod;e=read()%Mod;f=read()%Mod;for(int i=1;i<=n;i++){s[i]=(((a%mod)*(s[i-1]%mod)*(s[i-1])%mod)%mod+((b%mod)*(s[i-1]%mod))%mod+c%mod)%mod;ans|=ans<<s[i];}for(int i=1;i<=m;i++){y=((((y%Mod)*(y%Mod))%Mod*(d%Mod))%Mod+((y%Mod)*(e%Mod))%Mod+f%Mod)%Mod;if(ans[y])printf("yes\n");elseprintf("no\n");}}
}
int qty=yjn();
int main(){;}
