Problems: 给你一个数n和代价分别为a, b, c、数量不限的1, 2, 3，求将n凑成4的倍数的最小代价
Analysis:
cj:取个模随便凑一凑就好
Tags: Implementation
1 #define PRON "pa"
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7 typedef long long ll;
8
9 inline ll _min(ll a, ll b, ll c){
10 return min(a, min(b, c));
11 }
12
13 int main(){
14 ll n, a, b, c;
15 cin >> n >> a >> b >> c;
16
17 ll r = n % 4, ans = 0;
18 if (r == 1)
19 ans = _min(3 * a, a + b, c);
20 if (r == 2)
21 ans = _min(2 * a, b, c * 2);
22 if (r == 3)
23 ans = _min(a, b * 2 + c, c * 3);
24
25 cout << ans;
26 }
code by cj
Problems: 一个数列，每次操作选择一个区间[l, r]；若a[i]被选择了b[i]次，则Alyona的幸福值增加a[i] * b[i]；保留一些操作使得最后幸福值最高。
Analysis:
cj:对于每次操作，对于幸福值的贡献是sum{a[l], a[l + 1], ... , a[r]}，只要贡献是正的的操作就保留
Tags: Implementation
1 #define PRON "pb"
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 #define inf 0x3f3f3f3f
7 using namespace std;
8 typedef long long ll;
9
10 const int maxn = 100 + 10;
11
12 int n, m, a[maxn];
13
14 int main(){
15 #ifndef ONLINE_JUDGE
16 freopen(PRON ".in", "r", stdin);
17 #endif
18
19 cin >> n >> m;
20 for (int i = 1; i <= n; i ++)
21 cin >> a[i];
22
23 int ans = 0, temp, l, r;
24 while (m --){
25 cin >> l >> r;
26
27 temp = 0;
28 for (int i = l; i <= r; i ++)
29 temp += a[i];
30 if (temp > 0)
31 ans += temp;
32 }
33
34 cout << ans << endl;
35 }
code by cj
Problems: 给定m个区间，构造一个长度为n的数列，使得每个区间的mex的最小值最大。
Analysis:
cj:首先这个mex的最小值的最大值显然是min_mex = min{r[i] - l[i] + 1}；最简单的构造办法就是用{1, 2, 3 ..., min_mex - 1}来填充这个数列。
Tags: Constructive Algorithms
1 #define PRON "pc"
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 #define inf 0x3f3f3f3f
7 using namespace std;
8 typedef long long ll;
9
10 const int maxn = 2e5 + 10;
11
12 int n, m;
13 pair<int, int> a[maxn];
14
15 bool cmp(pair<int, int> A, pair<int, int> B){
16 return A.second - A.first < B.second - B.first;
17 }
18
19 int main(){
20 #ifndef ONLINE_JUDGE
21 freopen(PRON ".in", "r", stdin);
22 #endif
23
24 cin >> n >> m;
25
26 int mex = inf;
27 for (int i = 0; i < m; i ++){
28 cin >> a[i].first >> a[i].second;
29 mex = min(mex, 1 + a[i].second - a[i].first);
30 }
31
32 cout << mex << endl;
33
34 for (int i = 1; i <= n; i ++)
35 cout << i % mex << " ";
36 cout << endl;
37 }
code by cj
Problems: 一棵以1为根的树和各边的长度，给定每个点有一个值a。当dis(v, u) < a[u]并且u在v的子树中就认为u控制了v。问每个点控制了多少个点。
Analysis:
cj:dis(v, u) < a[u]可以变成d[u] - d[v] < a[u]，即：d[v] > d[u] - a[u]，其中d[i]表示dis(1, i)。用vector来维护dfs序，显然d是递增的，所以每次二分查找这个v，使得v以上的所以点都不能控制u，以下的点都可以控制u。用数组c来记录答案，初值赋1，装作能被控制的样子，对于一个点x，c[x] = sum{c[i]}，i是x的子树中的点。所以当查找到v之后，将这条链上v的亲爸爸的c减1，这样的话更新答案的时候就不会加上这个点及其以上的点。最后输出c[i] - 1即可。
Tags: binary search, dfs
1 #define PRON "pd"
2 #include <vector>
3 #include <cstdio>
4 #include <cstring>
5 #include <iostream>
6 #include <algorithm>
7 #define inf 0x3f3f3f3f
8 using namespace std;
9 typedef long long ll;
10
11 const int maxn = 2e5 + 10;
12
13 struct Edge {
14 int to, d;
15 Edge * next;
16 } e[maxn], * head[maxn];
17
18 vector<int> q;
19 vector<ll> dis;
20 int n, ne = 0, a[maxn], c[maxn];
21 ll d[maxn];
22
23 inline void add_edge(int f, int to, int d){
24 e[ne].to = to, e[ne].d = d;
25 e[ne].next = head[f];
26 head[f] = e + ne ++;
27 }
28
29 void dfs(int u){
30 for (Edge * p = head[u]; p; p = p -> next){
31 d[p -> to] = d[u] + (ll)p -> d;
32 dfs(p -> to);
33 }
34 }
35
36 void solve(int u){
37 c[u] = 1;
38 q.push_back(u);
39 dis.push_back(d[u]);
40
41 int anc = lower_bound(dis.begin(), dis.end(), d[u] - (ll)a[u]) - dis.begin();
42
43 if (anc >= 1)
44 c[q[anc - 1]] --;
45
46 for (Edge * p = head[u]; p; p = p -> next){
47 solve(p -> to);
48 c[u] += c[p -> to];
49 }
50
51 q.pop_back();
52 dis.pop_back();
53 }
54
55 int main(){
56 #ifndef ONLINE_JUDGE
57 freopen(PRON ".in", "r", stdin);
58 #endif
59
60 scanf("%d", &n);
61 for (int i = 1; i <= n; i ++)
62 scanf("%d", &a[i]);
63
64 int fa, dis;
65 for (int i = 2; i <= n; i ++){
66 scanf("%d %d", &fa, &dis);
67 add_edge(fa, i, dis);
68 }
69
70 d[1] = 0;
71 dfs(1);
72 solve(1);
73
74 for (int i = 1; i <= n; i ++)
75 printf("%d ", c[i] - 1);
76 printf("\n");
77 }
code by cj
Problems:给一个数列a，每次操作让[l, r]上的数增加的，每次操作后求满足al?<?al?+?1?<?al?+?2?<?...?<?ak?>?ak?+?1?>?ak?+?2?>?...?>?ar的最大区间长度。
Analysis:
cj:把数列换成差，这样的话每次区间修改操作，因为区间中的差是不变的，只有diff(arr[l - 1], arr[l])增加了d，diff(arr[r], arr[r + 1])减少了d，所以就变成了修改两个点的操作。用线段树维护一下就好了。不过要注意一下int要爆炸...并且要特判n = 1的情况。
Tags: Data structure
1 #define PRON "740e"
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 #define inf 0x3f3f3f3f
7 #define ls root << 1
8 #define rs root << 1 | 1
9 #define mid ((l + r) >> 1)
10 #ifdef UNIX
11 #define LL "%lld"
12 #else
13 #define LL "%I64d"
14 #endif
15 using namespace std;
16 typedef long long ll;
17
18 const int maxn = 3e5 + 10;
19
20 struct node {
21 int lans, rans, ans;
22 } t[maxn << 2];
23
24 int n, m;
25 ll arr[maxn], a[maxn];
26
27 void push_up(int l, int r, int root){
28 t[root].ans = max(t[ls].ans, t[rs].ans);
29 t[root].lans = t[ls].lans;
30 t[root].rans = t[rs].rans;
31
32 if ((a[mid] > 0 && a[mid + 1] != 0) || (a[mid] < 0 && a[mid + 1] < 0)){
33 t[root].ans = max(t[root].ans, t[ls].rans + t[rs].lans);
34 if (mid - l + 1 == t[ls].lans)
35 t[root].lans = t[ls].lans + t[rs].lans;
36 if (r - mid == t[rs].rans)
37 t[root].rans = t[ls].rans + t[rs].rans;
38 }
39 }
40
41 void build_tree(int l, int r, int root){
42 if (l == r){
43 t[root].ans = t[root].lans = t[root].rans = a[l] ? 1 : 0;
44 return;
45 }
46
47 build_tree(l, mid, ls);
48 build_tree(mid + 1, r, rs);
49
50 push_up(l, r, root);
51 }
52
53 void update(int l, int r, int pos, int root){
54 if (l == r && l == pos){
55 t[root].ans = t[root].lans = t[root].rans = a[l] ? 1 : 0;
56 return;
57 }
58
59 if (l <= pos && pos <= mid)
60 update(l, mid, pos, ls);
61 else
62 update(mid + 1, r, pos, rs);
63
64 push_up(l, r, root);
65 }
66
67 int main(){
68 #ifndef ONLINE_JUDGE
69 freopen(PRON ".in", "r", stdin);
70 #endif
71
72 scanf("%d", &n);
73 for (int i = 0; i < n; i ++){
74 scanf(LL, &arr[i]);
75 if (i)
76 a[i] = arr[i] - arr[i - 1];
77 }
78
79 if (n != 1)
80 build_tree(1, n - 1, 1);
81
82 int l, r, d;
83 scanf("%d", &m);
84 while (m --){
85 scanf("%d %d %d", &l, &r, &d);
86 if (n == 1){
87 printf("1\n");
88 continue;
89 }
90 if (l > 1){
91 a[l - 1] += d;
92 update(1, n - 1, l - 1, 1);
93 }
94 if (r < n){
95 a[r] -= d;
96 update(1, n - 1, r, 1);
97 }
98
99 printf("%d\n", t[1].ans + 1);
100 }
101 }
code by cj
时间： 2024-10-13 17:20:29