bitset+DFS序+线段树 Codeforces633G Yash And Trees

传送门：点击打开链接

题意：给你一棵树，根节点为1

有2种操作，第一种是给u节点所在的子树的所有节点的权值+x

第二种是询问，假设v是子树u中的节点，有多少种质数满足av = p + m·k

其中p<m

思路：首先用DFS序，把树弄成线段树来表示，这种做法十分常见就不多讲了

因为m<=1000,我们用bitset保存av%m的值

每次对子树增加权值的时候，只需要修改懒惰标记，来记录增加的大小

然后直接把bitset利用位运算来完成循环移动就行了。

这道题只要能熟练运用bitset，对DFS序的线段树有一定的了解，应该是没很大问题的。

#include<map>#include<set>#include<cmath>#include<ctime>#include<stack>#include<queue>#include<cstdio>#include<cctype>#include<bitset>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define fuck(x) cout<<"["<<x<<"]"#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;const int MX = 1e5 + 5;const int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;const int M = 1005;#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1int n, m, Q;int DL[MX], DR[MX], DFN;int A[MX], B[MX], sum[MX << 2];bitset<M>stu[MX << 2], temp, pr;int Head[MX], erear;struct Edge {    int v, nxt;} E[MX * 2];void edge_init() {    erear = 0;    memset(Head, -1, sizeof(Head));}void edge_add(int u, int v) {    E[erear].v = v;    E[erear].nxt = Head[u];    Head[u] = erear++;}bool prime_is(int x) {    for(int i = 2; i * i <= x; i++) {        if(x % i == 0) return false;    }    return true;}void prime_init() {    pr = 0;    for(int i = 2; i < m; i++) {        if(prime_is(i)) pr.set(i);    }}void DFS(int u, int f) {    DL[u] = ++DFN;    B[DFN] = A[u];    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(v == f) continue;        DFS(v, u);    }    DR[u] = DFN;}void push_up(int rt) {    stu[rt] = stu[rt << 1] | stu[rt << 1 | 1];}void func_right(int rt, int x) {    stu[rt] = (stu[rt] << x) | (stu[rt] >> (m - x));}void push_down(int rt) {    if(sum[rt]) {        sum[rt << 1] += sum[rt]; sum[rt << 1] %= m;        sum[rt << 1 | 1] += sum[rt]; sum[rt << 1 | 1] %= m;        func_right(rt << 1, sum[rt]);        func_right(rt << 1 | 1, sum[rt]);        sum[rt] = 0;    }}void build(int l, int r, int rt) {    sum[rt] = 0;    if(l == r) {        stu[rt].reset();        stu[rt].set(B[l] % m);        return;    }    int m = (l + r) >> 1;    build(lson); build(rson);    push_up(rt);}void add(int L, int R, int x, int l, int r, int rt) {    if(L <= l && r <= R) {        sum[rt] = (sum[rt] + x) % m;        func_right(rt, x);        return;    }    int m = (l + r) >> 1;    push_down(rt);    if(L <= m) add(L, R, x, lson);    if(R > m) add(L, R, x, rson);    push_up(rt);}bitset<M> query(int L, int R, int l, int r, int rt) {    if(L <= l && r <= R) {        return stu[rt];    }    int m = (l + r) >> 1;    push_down(rt);    bitset<M> ret;    if(L <= m) ret |= query(L, R, lson);    if(R > m) ret |= query(L, R, rson);    return ret;}int main() {    //FIN;    while(~scanf("%d%d", &n, &m)) {        DFN = 0;        edge_init(); prime_init();        for(int i = 1; i <= n; i++) {            scanf("%d", &A[i]);        }        for(int i = 1; i <= n - 1; i++) {            int u, v;            scanf("%d%d", &u, &v);            edge_add(u, v);            edge_add(v, u);        }        DFS(1, -1);        build(1, n, 1);        scanf("%d", &Q);        while(Q--) {            int op, a, b;            scanf("%d%d", &op, &a);            if(op == 1) {                scanf("%d", &b);                add(DL[a], DR[a], b % m, 1, n, 1);            } else {                temp = query(DL[a], DR[a], 1, n, 1);                printf("%d\n", (int)((temp & pr).count()));            }        }    }}