AtCoder Beginner Contest 255 Solution更好的阅读体验戳此进入题面链接题面 Luogu 链接abcd 跳了[ABC255E] Lucky Numbers题面SolutionCode[ABC255F] Pre-order and In-order题面SolutionCode[ABC255G] Constrained Nim题面SolutionCode[ABC255Ex] Range Harvest Query题面SolutionCodeUPD
给定长度为
首先不难发现,对于序列
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601
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11using namespace std;
12
13mt19937 rnd(random_device{}());
14int rndd(int l, int r){return rnd() % (r - l + 1) + l;}
15bool rnddd(int x){return rndd(1, 100) <= x;}
16
17typedef unsigned int uint;
18typedef unsigned long long unll;
19typedef long long ll;
20typedef long double ld;
21
22template < typename T = int >
23inline T read(void);
24
25int N, M;
26int S[110000];
27int lucky[20];
28ll sum[110000];
29int ans(-1);
30unordered_map < ll, int > mp;
31
32int main(){
33 N = read(), M = read();
34 for(int i = 1; i <= N - 1; ++i)
35 S[i] = read(), sum[i] = sum[i - 1] + (i & 1 ? 1 : -1) * S[i];
36 for(int i = 1; i <= M; ++i)lucky[i] = read();
37 for(int i = 1; i <= N; ++i)
38 for(int j = 1; j <= M; ++j)
39 mp[(i & 1 ? 1 : -1) * lucky[j] + sum[i - 1]]++;
40 for(auto v : mp)ans = max(ans, v.second);
41 printf("%d\n", ans);
42 fprintf(stderr, "Time: %.6lf\n", (double)clock() / CLOCKS_PER_SEC);
43 return 0;
44}
45
46template < typename T >
47inline T read(void){
48 T ret(0);
49 int flag(1);
50 char c = getchar();
51 while(c != '-' && !isdigit(c))c = getchar();
52 if(c == '-')flag = -1, c = getchar();
53 while(isdigit(c)){
54 ret *= 10;
55 ret += int(c - '0');
56 c = getchar();
57 }
58 ret *= flag;
59 return ret;
60}
给定一棵二叉树的先序遍历和中序遍历,请构造一棵以 -1
。
也是一道水题,考虑先序和中序的意义即可。
众所周知,先序遍历的顺序是根、左子树、右子树。中序遍历是左子树、根、右子树。
于是不难发现,在当前的先序遍历中取第一个数即为当前的根,然后在中序遍历中找到根的位置,其左侧即为整个左子树,右侧为整个右子树。于是不难想到 dfs 即可,参数维护当前的整个子树属于先序遍历的
考虑无解的情况,要么先序遍历的第一个值不为
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11using namespace std;
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13mt19937 rnd(random_device{}());
14int rndd(int l, int r){return rnd() % (r - l + 1) + l;}
15bool rnddd(int x){return rndd(1, 100) <= x;}
16
17typedef unsigned int uint;
18typedef unsigned long long unll;
19typedef long long ll;
20typedef long double ld;
21
22template < typename T = int >
23inline T read(void);
24
25int N;
26int Pre[210000], In[210000];
27int posP[210000], posI[210000];
28pair < int, int > son[210000];
29
30int dfs(int lp = 1, int rp = N, int li = 1, int ri = N){
31 // printf("In dfs(%d ~ %d, %d ~ %d)\n", lp, rp, li, ri);
32 if(lp > rp)return 0;
33 int rt = Pre[lp];
34 if(posI[rt] < li || posI[rt] > ri)puts("-1"), exit(0);
35 if(lp == rp)return rt;
36 int lsiz = (posI[rt] - 1) - li + 1;
37 son[rt].first = dfs(lp + 1, lp + lsiz, li, posI[rt] - 1);
38 son[rt].second = dfs(lp + lsiz + 1, rp, posI[rt] + 1, ri);
39 return rt;
40}
41
42int main(){
43 N = read();
44 for(int i = 1; i <= N; ++i)posP[Pre[i] = read()] = i;
45 for(int i = 1; i <= N; ++i)posI[In[i] = read()] = i;
46 if(Pre[1] != 1)puts("-1"), exit(0);
47 dfs();
48 for(int i = 1; i <= N; ++i)printf("%d %d\n", son[i].first, son[i].second);
49 fprintf(stderr, "Time: %.6lf\n", (double)clock() / CLOCKS_PER_SEC);
50 return 0;
51}
52
53template < typename T >
54inline T read(void){
55 T ret(0);
56 int flag(1);
57 char c = getchar();
58 while(c != '-' && !isdigit(c))c = getchar();
59 if(c == '-')flag = -1, c = getchar();
60 while(isdigit(c)){
61 ret *= 10;
62 ret += int(c - '0');
63 c = getchar();
64 }
65 ret *= flag;
66 return ret;
67}
一般 Nim 游戏基础上增加
显然博弈论,考虑 SG函数。Nim 游戏标准套路,对于多个石子堆求出每个石子堆石子数
本题的区别即为 map
记录转折点,每次查询对应的所在位置然后增加对应数量的
具体实现过程也不难理解,开个 map
里套 basic_string
维护对应
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11using namespace std;
12
13mt19937 rnd(random_device{}());
14int rndd(int l, int r){return rnd() % (r - l + 1) + l;}
15bool rnddd(int x){return rndd(1, 100) <= x;}
16
17typedef unsigned int uint;
18typedef unsigned long long unll;
19typedef long long ll;
20typedef long double ld;
21
22template < typename T = int >
23inline T read(void);
24
25int N, M;
26ll A[210000];
27ll oplus(0);
28map < ll, basic_string < ll > > rest;
29map < ll, ll > SG, repeat;
30
31ll CalSG(ll x){
32 auto sp = *prev(SG.upper_bound(x));
33 return sp.second + (x - sp.first);
34}
35
36int main(){
37 N = read(), M = read(); SG.insert({0, 0});
38 for(int i = 1; i <= N; ++i)A[i] = read < ll >();
39 for(int i = 1; i <= M; ++i){
40 ll p = read < ll >(), v = read < ll >();
41 rest[p] += p - v;
42 }
43 ll preMx(-1);
44 for(auto &mp : rest){
45 preMx = max(preMx, CalSG(mp.first - 1));
46 map < ll, ll > tmp;
47 for(auto val : mp.second)tmp[CalSG(val)]++;
48 for(auto cur : tmp){
49 if(cur.second >= repeat[cur.first] + 1){
50 repeat[cur.first]++;
51 SG[mp.first] = cur.first;
52 break;
53 }
54 }preMx = max(preMx, CalSG(mp.first));
55 SG[mp.first + 1] = preMx + 1;
56 }
57 for(int i = 1; i <= N; ++i)oplus ^= CalSG(A[i]);
58 printf("%s\n", oplus ? "Takahashi" : "Aoki");
59 fprintf(stderr, "Time: %.6lf\n", (double)clock() / CLOCKS_PER_SEC);
60 return 0;
61}
62
63template < typename T >
64inline T read(void){
65 T ret(0);
66 int flag(1);
67 char c = getchar();
68 while(c != '-' && !isdigit(c))c = getchar();
69 if(c == '-')flag = -1, c = getchar();
70 while(isdigit(c)){
71 ret *= 10;
72 ret += int(c - '0');
73 c = getchar();
74 }
75 ret *= flag;
76 return ret;
77}
给定一片
给出
ODT 随便搞一下即可,没有任何难度,处理一下细节即可。一下子居然没看出来,日常身败名裂。
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11using namespace std;
12
13mt19937 rnd(random_device{}());
14int rndd(int l, int r){return rnd() % (r - l + 1) + l;}
15bool rnddd(int x){return rndd(1, 100) <= x;}
16
17typedef unsigned int uint;
18typedef unsigned long long unll;
19typedef long long ll;
20typedef long double ld;
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23
24template < typename T = int >
25inline T read(void);
26
27struct Node{
28 ll l, r;
29 mutable ll lst;
30 friend const bool operator < (const Node &a, const Node &b){
31 return a.l < b.l;
32 }
33};
34
35class ODT{
36private:
37 set < Node > tr;
38public:
39 auto Insert(Node p){return tr.insert(p);}
40 auto Split(ll p){
41 auto it = tr.lower_bound(Node{p});
42 if(it != tr.end() && it->l == p)return it;
43 advance(it, -1);
44 ll l = it->l, r = it->r, lst = it->lst;
45 tr.erase(it);
46 Insert(Node{l, p - 1, lst});
47 return Insert(Node{p, r, lst}).first;
48 }
49 ll Assign(ll l, ll r, ll days){
50 ll ret(0);
51 auto itR = Split(r + 1), itL = Split(l);
52 for(auto it = itL; it != itR; ++it)
53 (ret += ((__int128_t)(it->l + it->r) * (it->r - it->l + 1) / 2ll) % MOD * (days - it->lst) % MOD) %= MOD;
54 tr.erase(itL, itR);
55 Insert(Node{l, r, days});
56 return ret;
57 }
58}odt;
59
60ll N; int Q;
61
62int main(){
63 N = read < ll >(), Q = read();
64 odt.Insert(Node{1, N, 0});
65 while(Q--){
66 ll D = read < ll >(), L = read < ll >(), R = read < ll >();
67 printf("%lld\n", odt.Assign(L, R, D));
68 }
69 fprintf(stderr, "Time: %.6lf\n", (double)clock() / CLOCKS_PER_SEC);
70 return 0;
71}
72
73template < typename T >
74inline T read(void){
75 T ret(0);
76 int flag(1);
77 char c = getchar();
78 while(c != '-' && !isdigit(c))c = getchar();
79 if(c == '-')flag = -1, c = getchar();
80 while(isdigit(c)){
81 ret *= 10;
82 ret += int(c - '0');
83 c = getchar();
84 }
85 ret *= flag;
86 return ret;
87}
update-2022_12_07 初稿