problem
analysis
正解分块大法好
考虑把原数组分成 块,每块大小为 ,对每一块做前缀积和后缀积
对于每一位 ,如果它是一个块的开头,答案异或上这块后缀积,否则异或上 位的前缀积
注意这题卡空间,于是我被 卡出屎
code
#include<stdio.h>
#define MAXN 20000005
#define ll long long
#define fo(i,a,b) for (register int i=a;i<=b;i++)
#define fd(i,a,b) for (register int i=a;i>=b;i--)
using namespace std;
int a[MAXN],b[MAXN],c[MAXN];
int n,k,p,A,B,C,D,ans;
long long temp;
int read()
{
int x=0,f=1;
char ch=getchar();
while (ch<'0' || '9'<ch)
{
if (ch=='-')f=-1;
ch=getchar();
}
while ('0'<=ch && ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
int main()
{
//freopen("readin.txt","r",stdin);
freopen("range.in","r",stdin);
freopen("range.out","w",stdout);
n=read(),k=read(),p=read(),A=read(),B=read(),C=read(),D=read();
a[1]=A;
fo(i,2,n)a[i]=((ll)(a[i-1])*(ll)(B)+(ll)C)%D;
int x=0,y=0;
do
{
x=y+1,y=(x+k-1>n?n:x+k-1);
b[x]=a[x],c[y]=a[y];
fo(l,x+1,y)
{
temp=(ll)(b[l-1])*(ll)(a[l])%p;
b[l]=temp;
}
fd(l,y-1,x)
{
temp=(ll)(c[l+1])*(ll)(a[l])%p;
c[l]=temp;
}
}while (y!=n);
fo(i,1,n-k+1)
{
temp=(ll)(c[i])*(ll)(b[i+k-1])%p;
ans^=(i%k==1?c[i]:temp);
}
printf("%d\n",ans);
return 0;
}