zoj3886--Nico Number（素数筛+线段树）

Posted 2020-09-21 wzjhoutai

Nico Number

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Time Limit: 2 Seconds      Memory Limit: 262144 KB

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Kousaka Honoka and Minami Kotori are playing a game about a secret of Yazawa Nico.

When the game starts, Kousaka Honoka will give Minami Kotori an array A of N non-negative integers. There is a special kind of number in the array, which is called NicoNico-number. We call a integer x is a NicoNico-number, if all integers(no more than x) that is coprime with x could form an Arithmetic Sequence.

Then Minami Kotori will choose some consecutive part of the array A, wondering the number of NicoNico-number in this part. What‘s more, Kousaka Honoka sometimes modify the value of some consecutive elements in the array. Now it is your task to simulate this game!

Input

There are multiple test cases in input, you should process to the end of file.

For each case, the first line is an integer N, the number of elements in the array described above. Then second line contains N integers no greater than 10⁷, the elements of the array initially.(1 <= N <= 100,000)

The third line is a integer T, the number of the operations of the game. Each line of the following T lines is in one of the following formats.(1 <= T <= 100,000)

"1 L R" : Minami Kotori will chooses the consecutive part of the array from the Lth to Rth element inclusive. (1 <= L <= R <= N)

"2 L R v" : Kousaka Honoka will change the value of the pth element A[p] in the array to A[p]%v for all L <= p <= R.(1 <= L <= R <= N, 1 <= v <= 10⁷)

"3 p x" : Kousaka Honoka will change the value of the p th element A[p] to x.(1 <= p <= N, 1 <= x <= 10⁷)

Output

Each time when Minami Kotori chooses some part of the array, you should output a line, the number of NicoNico-number in that part.

Sample Input

3
4 6 9
6
1 1 3
1 3 3
2 1 1 10
1 1 3
3 2 4
1 1 3

Sample Output

2
0
2
2

Hint

4 is a NicoNico-number because only 1 and 3 is coprime with 4 among the integers no greater than 4, and could form an Arithmetic Sequence {1,3}.

题目大意：定义一个NicoNico-number，假设x是NicoNico-number，那么全部小于x的且与x互质的整数是一个等差数列，初始给出n个数字的数组，三种操作：

1 l r 问在[l,r]内有多少个NicoNico-number数

2 l r v 对于[l,r]内的数所有对v取余

3 k x 将第k个数换为x

对每一次询问做出输出。

1、首先写一个找规律的，发现NicoNico-number是有三种组成的第一种是素数，另外一种是2的x次幂，第三种是6

2、那么能够建一个数组。直接标记某个数是不是NicoNico-number

3、使用线段树维护一段区间的NicoNico-number个数，然后能够进行对某一个数的改动，和对一个区间的查询。

对于另外一种操作。我们要知道对于一个数x取余操作。最多会运行log(x)次。由于每次取余至少数值会降低一半。所以对于每一个数来说最多会有log(x)次操作，之后会由于v大于当前值，而不用运行操作。既然取余的次数不多，那么就能够对区域操作进行暴力，维护一段区间的最大值，假设最大值小于v，那么这一段不用更新，否则就遍历的最低层进行取余。

4、对于n个数来说，查找到一个数须要log(n)，一个数最多会被改动log(x)次，所以总的时间不会超过n*log(n)*log(x)。

#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <cmath>
#include <map>
#include <stack>
#include <algorithm>
using namespace std ;
#define LL __int64
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define root 1,n,1
#define int_rt int l,int r,int rt
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
const int mod = 1e9+7 ;
const double eqs = 1e-9 ;
int cl[400000] , num[400000] ;
int a[10000005] , check[10000005] ;
int tot ;
void init() {
    memset(check,-1,sizeof(check)) ;
    tot = 0 ;
    for(int i = 2 ; i <= 10000000 ; i++) {
        if( check[i] == -1 ){
            a[tot++] = i ;
            check[i] = 1 ;
        }
        for(int j = 0 ; j < tot ; j++) {
            if( i*a[j] >= 10000000 ) break ;
            check[i*a[j]] = 0 ;
            if( i%a[j] == 0 ) break ;
        }
    }
    check[0] = check[1] = check[6] = 1 ;
    for(int i = 2 ; i <= 10000000 ; i *= 2)
        check[i] = 1 ;
}
void push_up(int rt) {
    cl[rt] = max(cl[rt<<1],cl[rt<<1|1]) ;
    num[rt] = num[rt<<1]+num[rt<<1|1] ;
}
void create(int_rt) {
    cl[rt] = num[rt] = 0 ;
    if( l == r ) {
        scanf("%d", &cl[rt]) ;
        if( check[ cl[rt] ] == 1 ) num[rt] = 1 ;
        return ;
    }
    create(lson) ;
    create(rson) ;
    push_up(rt) ;
}
void update1(int ll,int rr,int v,int_rt) {
    if( ll > r || rr < l ) return ;
    if( cl[rt] < v ) return ;
    if( l == r ) {
        cl[rt] %= v ;
        if( check[ cl[rt] ] == 1 ) num[rt] = 1 ;
        else num[rt] = 0 ;
        return ;
    }
    update1(ll,rr,v,lson) ;
    update1(ll,rr,v,rson) ;
    push_up(rt) ;
}
void update2(int k,int x,int_rt) {
    if( l == r && l == k ) {
        cl[rt] = x ;
        if( check[ cl[rt] ] == 1 ) num[rt] = 1 ;
        else num[rt] = 0 ;
        return ;
    }
    int mid = (l+r)/2 ;
    if(k <= mid) update2(k,x,lson) ;
    else update2(k,x,rson) ;
    push_up(rt) ;
}
int query(int ll,int rr,int_rt) {
    if( ll > r || rr < l ) return 0 ;
    if( ll <= l && rr >= r ) return num[rt] ;
    return query(ll,rr,lson) + query(ll,rr,rson) ;
}
int main() {
    int n , m , i , k , l , r , v , x ;
    init() ;
    while( scanf("%d", &n) !=EOF ) {
        create(root) ;
        scanf("%d", &m) ;
        while( m-- ) {
            scanf("%d", &k) ;
            if( k == 1 ) {
                scanf("%d %d", &l, &r) ;
                printf("%d\n", query(l,r,root)) ;
            }
            else if( k == 2 ) {
                scanf("%d %d %d", &l, &r, &v) ;
                update1(l,r,v,root) ;
            }
            else {
                scanf("%d %d", &i, &x) ;
                update2(i,x,root) ;
            }
        }
    }
    return 0 ;
}