Codeforces Round #169 (Div. 2)补题记录

前三题没什么可说的

CF - 276C.Little Girl and Maximum Sum

记录每个索引询问的次数，将数组内的数从大到小排序，询问次数从大到小排序，相乘求和即可。

CF - 276D.Little Girl and Maximum XOR

求$[l,r]$区间内两个数异或的最大值。

我们考虑$l$和$r$异或所得的数$x$，我们的答案就是从这个数$x$的最高位$1$开始，后面所有位全部置为$1$。

注意$long~long$

#if __cplusplus >= 201103L
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC optimize(3, "Ofast", "inline")
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cmath>
#include <iostream>
#include <map>
#include <string.h>
#include <vector>
#endif

using namespace std;
#define inf __INT_MAX__
#define enf INT_MIN
#define INF LLONG_MAX
#define ENF LLONG_MIN

const int MAXN = 2e5 + 10;
const double pi = acos(-1.0);
const double eps = 1e-6;
typedef long long ll;
typedef unsigned long long ull;
#define zhengfu(x) ((x > eps) - (x < -eps))

int main() {
    ll l, r;
    cin >> l >> r;
    if (l == r)
        cout << 0 << endl;
    else {
        ll tmp = l ^ r;
        for (ll i = 63; i >= 0; i--)
            if (tmp & (1LL << i)) {
                cout << (1LL << (i + 1)) - 1LL << endl;
                break;
            }
    }
}

CF - 276E.Little Girl and Problem on Trees

给你一棵树，其中，除了根节点$1$以外，其他的节点度数不大于2。

也就是说，如果把$1$从中删除，那么就会变成一条条链。

两个操作，一个操作是对于节点$v$，令与其距离不超过$d$的节点加$x$，包括自己。

另一个是询问节点$v$上的加和是多少。

考虑题面：

​ 区间更新，单点查询$\longrightarrow$显然线段树$or$树状数组

​ 考虑区间更新：

• 当距离$d$只局限于节点$v$所在的链上，即$depth[v]-d$不包括根节点$1$时

  只需要对于该链的节点进行简单的区间更新即可。

• 当距离$d$不局限于$v$所在链时，显然，更新包括了根节点$1$

  • 那么， 我们可以根据树的深度建一个树状数组，我们只需要更新$[1,d-depth[v]]$之间的节点即可

  • 所以，我们的更新过程就成了这样：

    • 首先在链上局部更新新$x$，范围是$[depth[v]-d,depth[v]+d]$
    • 然后在深度树状数组更新$x$，范围是$[1,d-depth[v]]$
    • 然后在链上局部更新$-x$，范围是$[1,d-depth[v]]$

​ 再考虑单点查询，答案显然就变成了，深度树状数组$+$链上树状数组的和。

#if __cplusplus >= 201103L
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC optimize(3, "Ofast", "inline")
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cmath>
#include <iostream>
#include <map>
#include <string.h>
#include <vector>
#endif

using namespace std;
#define inf __INT_MAX__
#define enf INT_MIN
#define INF LLONG_MAX
#define ENF LLONG_MIN

const int MAXN = 2e5 + 10;
const double pi = acos(-1.0);
const double eps = 1e-6;
typedef long long ll;
typedef unsigned long long ull;
#define zhengfu(x) ((x > eps) - (x < -eps))

/*
每条链一个树状数组,记录链上不过根的更新
一个总树状数组记录过根的深度的更新
*/
vector<int> e[MAXN];
struct node {
    vector<int> nd;
    int n;
    void init() {
        n = nd.size() - 1;
    }
    inline int lowbit(int x) //核心操作,取x这个数的二进制最后一位1
    {
        return -x & x;
    }
    inline void update(int x, int k) //单点修改
    {
        for (; x <= n; x += lowbit(x))
            nd[x] += k;
    }
    void update(int l, int r, int k) {  //区间更新
        update(l, k);
        if (r < n)
            update(r + 1, -k);
    }
    inline int query(int x) //求x的前缀和
    {
        int sum(0);
        for (; x > 0; x -= lowbit(x))
            sum += nd[x];
        return sum;
    }
} tree[MAXN];
int depmax = enf;
int dp[MAXN];
int nm[MAXN];
void dfs(int anc, int x, int dep, int num) {
    tree[num].nd.push_back(0);
    dp[x] = dep, nm[x] = num;
    if (e[x].size() == 1) {
        depmax = max(dep, depmax);
        tree[num].init();
        return;
    }
    for (auto it : e[x])
        if (it != anc)
            dfs(x, it, dep + 1, num);
}
int main() {
    int n, q;
    cin >> n >> q;
    for (int i = 1, x, y; i < n; i++) {
        cin >> x >> y;
        e[x].push_back(y);
        e[y].push_back(x);
    }
    for (int i = 1; i <= e[1].size(); i++) {
        tree[i + 1].nd.push_back(0);
        tree[i + 1].nd.push_back(0);
        dfs(1, e[1][i - 1], 2, i + 1);
    }

    dp[1] = nm[1] = 1;
    for (int i = 0; i <= depmax; i++)
        tree[1].nd.push_back(0);
    tree[1].init();

    for (int i = 1, j, v, x, d; i <= q; i++) {
        cin >> j >> v;
        int pos = dp[v], tmp = nm[v];
        if (!j) {
            scanf("%d%d", &x, &d);
            tree[tmp].update(max(2, pos - d), pos + d, x);
            if (pos - d < 2) {
                int len = d - pos + 2;
                tree[tmp].update(2, len, -x);
                tree[1].update(1, len, x);
            }
        } else {
            if (v == 1)
                cout << tree[1].query(pos) << endl;
            else
                cout << tree[tmp].query(pos) + tree[1].query(pos) << endl;
        }
    }
}

六花保佑代码不$WA$

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