2021“MINIEYE杯”中国大学生算法设计超级联赛（8） - 1004（Counting Stars）

补题链接： ​​https://acm.hdu.edu.cn/showproblem.php?pid=7059​​

考点: ​​线段树​​

解题思路可参考​

但是与下面的代码不同的是: 下面代码考虑的是​​flag = true​​ 代表[l, r]区间内ai为0，即高位和低位都为0；

​​AC Code​​

#include <iostream>

using namespace std;

typedef long long ll;

const int mod = 998244353, N = 1e5 + 10;

int n, q;

// x 位代表a的二进制表示中最左边的1的位置
// lo: 记录a的前x-1位的大小
// hi: 记录a的x位的10进制表示的大小
ll lo[N], hi[N];

struct SegmentTree{

    int l, r;  // 区间
    ll sum1, sum2; // lo的区间和; hi的区间和
    bool flag;  // [l, r]区间每个ai都是0
    ll tag2; // ai + 2^k(相当于最高位*2)的懒标记; tag2的值代表[l, r]区间的每个hi需要乘多少

}tree[N << 2];

ll lowbit(ll x){
    return x & (-x);
}

void build(int node, int l, int r){

    // 初始化
    tree[node].l = l, tree[node].r = r;
    tree[node].flag = 0;
    tree[node].tag2 = 1;
    if(l == r){

        tree[node].sum1 = lo[l];
        tree[node].sum2 = hi[l];
        // 如果[l, r]的ai除最高位外,其余位都为0(sum1=0),flag=true;
        if(!tree[node].sum2) tree[node].flag = 1;
        return;
    }

    int mid = (l + r) >> 1;

    int left_node = node << 1;
    int right_node = node << 1 | 1;

    build(left_node, l, mid);
    build(right_node, mid + 1, r);

    tree[node].sum1 = (tree[left_node].sum1 + tree[right_node].sum1) % mod;
    tree[node].sum2 = (tree[left_node].sum2 + tree[right_node].sum2) % mod;

    // 只有两边都为0; 父节点才为0
    tree[node].flag = tree[left_node].flag & tree[right_node].flag;
}

// 关于pushDown操作: 当某操作向当前节点的子节点走时,一定要先执行pushDown操作
void pushDown(int node){

    int left_node = node << 1;
    int right_node = node << 1 | 1;

    tree[left_node].tag2 = tree[left_node].tag2 * tree[node].tag2 % mod;
    tree[right_node].tag2 = tree[right_node].tag2 * tree[node].tag2 % mod;

    tree[left_node].sum2 = tree[left_node].sum2 * tree[node].tag2 % mod;
    tree[right_node].sum2 = tree[right_node].sum2 * tree[node].tag2 % mod;

    tree[node].tag2 = 1;
}

// 暴力lowbit操作
void modify1(int node, int L, int R){

    if(tree[node].l == tree[node].r){

        // 如果当前ai的除了最高位都为0
        if(!tree[node].sum1){
            // 将 最高位 置0
            tree[node].sum2 = 0;
            tree[node].flag = 1;
            tree[node].tag2 = 1;
            return;
        }
        tree[node].sum1 -= lowbit(tree[node].sum1);

        return;
    }

    // 如果当前区间内所有位都是0
    // 所以tag2要变为1

    if(L <= tree[node].l && tree[node].r <= R && tree[node].flag){

        // tree[node].sum1 = tree[node].sum2 = 0;
        tree[node].tag2 = 1;
        return;
    }

    pushDown(node);

    int mid = (tree[node].l + tree[node].r) >> 1;
    int left_node = node << 1;
    int right_node = node << 1 | 1;

    if(L <= mid) modify1(left_node, L, R);
    if(R > mid) modify1(right_node, L, R);

    tree[node].sum1 = (tree[left_node].sum1 + tree[right_node].sum1) % mod;
    tree[node].sum2 = (tree[left_node].sum2 + tree[right_node].sum2) % mod;

    tree[node].flag = tree[left_node].flag & tree[right_node].flag;
}

// + 2^k操作, 相当于高位*2
void modify2(int node, int L, int R){

    if(L <= tree[node].l && tree[node].r <= R){

        tree[node].sum2 = tree[node].sum2 * 2 % mod;
        tree[node].tag2 = tree[node].tag2 * 2 % mod;
        return;
    }

    pushDown(node);

    int mid = (tree[node].l + tree[node].r) >> 1;
    int left_node = node << 1;
    int right_node = node << 1 | 1;

    if(L <= mid) modify2(left_node, L, R);
    if(R > mid) modify2(right_node, L, R);

    tree[node].sum1 = (tree[left_node].sum1 + tree[right_node].sum1) % mod;
    tree[node].sum2 = (tree[left_node].sum2 + tree[right_node].sum2) % mod;

    tree[node].flag = tree[left_node].flag & tree[right_node].flag;
}

// query
ll ask(int node, int L, int R){

    if(L <= tree[node].l && tree[node].r <= R){
        return (tree[node].sum1 + tree[node].sum2) % mod;
    }

    pushDown(node);

    ll res = 0;
    int mid = (tree[node].l + tree[node].r) >> 1;
    int left_node = node << 1;
    int right_node = node << 1 | 1;

    if(L <= mid) res += ask(left_node, L, R);
    if(R > mid) res = (res + ask(right_node, L, R)) % mod;

    return res;
}


int main(){

    int t;
    scanf("%d", &t);

    while (t--){

        scanf("%d", &n);

        ll a;
        for (int i = 1; i <= n; ++i) {

            scanf("%lld", &a);
            hi[i] = lo[i] = 0;  // 清空
            for (int j = 30; j > 0; j--) {

                // 计算a的lo值和hi值
                if((1ll << j) <= a){
                    hi[i] = (1ll << j);
                    lo[i] = a & (~hi[i]);
                    break;
                }
            }
        }

        // 建树
        build(1, 1, n);

        scanf("%d", &q);
        for (int i = 1; i <= q; ++i) {

            int opt, l, r;
            scanf("%d%d%d", &opt, &l, &r);
            if(opt == 1){
                printf("%lld\n", ask(1, l, r));
            }else if(opt == 2){
                modify1(1, l, r);
            }else{
                modify2(1, l, r);
            }
        }
    }

    return 0;
}