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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