ZOJ 3886 Nico number(线段树）

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 calledNicoNico-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

34 6 961 1 31 3 32 1 1 101 1 33 2 41 1 3

Sample Output

2022

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}.

解题思路：

只有6，2的整数次幂和质数满足要求，用线段树维护，区间取模的地方维护一个区间最大值，若区间最值小于mod则直接返回，不然暴力取模。

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>#include <queue>#include <set>#include <stack>#include <algorithm>#define LL long longusing namespace std;const int MAXN = 100000 + 10;const int MAXM = 10000000 + 10;int vis[MAXM], isprime[MAXM], prime[MAXM];void init(){    int m = (int)sqrt(MAXM + 0.5);    for(int i=2;i<=m;i++) if(!vis[i])    {        for(int j=i*i;j<=MAXM;j+=i)            vis[j] = 1;    }    vis[6] = vis[0] = 0;    for(int i=1;i<=MAXM;i<<=1) vis[i] = 0;}struct Tree{    int l, r;    int Max, sum;}tree[MAXN<<2];void push_up(int rt){    tree[rt].Max = max(tree[rt<<1].Max, tree[rt<<1|1].Max);    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;}void build(int l, int r, int rt){    tree[rt].l = l;    tree[rt].r = r;    tree[rt].Max = 0;    tree[rt].sum = 0;    if(l == r)    {        scanf("%d", &tree[rt].Max);        tree[rt].sum = 1 - vis[tree[rt].Max];        return ;    }    int mid = (l + r) >> 1;    build(l, mid, rt<<1);    build(mid + 1, r, rt<<1|1);    push_up(rt);}void update1(int p, int x, int rt){    if(tree[rt].l == tree[rt].r)    {        tree[rt].Max = x;        tree[rt].sum = 1 - vis[x];        return ;    }    int mid = (tree[rt].l + tree[rt].r) >> 1;    if(p <= mid) update1(p, x, rt<<1);    else update1(p, x, rt<<1|1);    push_up(rt);}void update2(int l, int r, int x, int rt){    if(tree[rt].l >= l && tree[rt].r <= r)    {        if(tree[rt].Max < x) return ;    }    if(tree[rt].l == tree[rt].r)    {        tree[rt].Max %= x;        tree[rt].sum = 1 - vis[tree[rt].Max];        return ;    }    int mid = (tree[rt].l + tree[rt].r) >> 1;    if(l <= mid) update2(l, r, x, rt<<1);    if(r > mid) update2(l, r, x, rt<<1|1);    push_up(rt);}int query(int l, int r, int rt){    //cout << l << ' ' << r << ' ' << tree[rt].l << ' ' << tree[rt].r << endl;    if(tree[rt].l >= l && tree[rt].r <= r)    {        return tree[rt].sum;    }    int mid = (tree[rt].l + tree[rt].r) >> 1;    int ans = 0;    if(l <= mid) ans += query(l, r, rt<<1);    if(r > mid) ans += query(l, r, rt<<1|1);    return ans;}int main(){    init();    int n, Q;    while(scanf("%d", &n)!=EOF)    {        build(1, n, 1);        scanf("%d", &Q);        int op, l, r;        while(Q--)        {            scanf("%d", &op);            if(op == 1)            {                scanf("%d%d", &l, &r);                printf("%d\n", query(l, r, 1));            }            else if(op == 2)            {                int x; scanf("%d%d%d", &l, &r, &x);                update2(l, r, x, 1);            }            else if(op == 3)            {                int p, x;                scanf("%d%d", &p, &x);                update1(p, x, 1);            }        }    }    return 0;}