【JZOJ100035】区间

problem

---

analysis

• 正解分块大法好

• 考虑把原数组分成$⌊{n\over k}⌋$块，每块大小为$k$，对每一块做前缀积和后缀积

• 对于每一位$i$，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$i+k-1$位的前缀积

• 注意这题卡空间，于是我被$long long$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;
}