Preface
不知道 VP 什么还是选了这赛季的外国场 UCUP，毕竟还是正规区域赛题目质量还是有保障的
就是整体难度相对偏低，签到和前期题比较多，同时算法题相对较少，不过这些问题也是老生常谈了
A. AIMPPZ
签到，枚举有几个 AI 即可
#include<cstdio>
#include<iostream>
using namespace std;
int n;
int main()
{scanf("%d",&n);int fact_5=0,tmp=n;while (tmp%5==0) ++fact_5,tmp/=5;int ans=n,times=0; tmp=1;for (int i=1;i<=fact_5;++i){tmp*=5;if (2*i>n/tmp) continue;if (n/tmp<ans) ans=n/tmp,times=i;}for (int i=1;i<=times;++i)printf("AI");for (int i=1;i<=ans-2*times;++i)printf("B");return 0;
}
B. Beats
考虑枚举一段前缀，钦定除了这段前缀外的后缀全部用第二种操作，此时只要求出需要多少次操作能给前缀排序
手玩一下会发现拿个队列维护一下可以不动的元素即可
#include<bits/stdc++.h>
using namespace std;const int N = 2e5+5;
int n, A[N];
int que[N], fr=0, ed=-1;
signed main() {ios::sync_with_stdio(0); cin.tie(0);cin >> n;int ans = n;for (int i=1; i<=n; ++i) cin >> A[i];for (int i=1; i<=n; ++i) {if (ed<fr || A[i]>que[ed]) que[++ed]=A[i];else {while (fr<=ed && A[i]>que[fr]) ++fr;}ans = min(ans, n-(ed-fr+1));}cout << ans << '\n';return 0;
}
C. Count Triangular Sequences
直接对 \(\{a_i\}\) 计数比较困难，我们考虑对其差分数组进行计数
令 \(d_i=a_i-(a_{i-1}+a_{i-2})\)，特别地，\(d_1=a_1,d_2=a_2-a_1\)
手玩一会我们会发现对于 \(i\ge 3\)，满足 \(a_i=\sum_{j=1}^n fib_j\times d_{i+1-j}\)，其中 \(fib\) 是以 \(1\) 开头的斐波那契数列
因此这个题本质上就变成一个类似于背包的问题，不难发现由于 \(a_i\le 200000\)，因此序列长度 \(\le 28\)，用一个类似于完全背包的东西计数即可
注意要特判序列开头两个元素相同的情形
#include<cstdio>
#include<iostream>
#include<cstring>
#define RI register int
#define CI const int&
using namespace std;
const int N=200005,mod=1e9+7,inv2=(mod+1)/2;
int n,fib[30],fact[30],f[30][N];
inline void inc(int& x,CI y)
{if ((x+=y)>=mod) x-=mod;
}
inline void dec(int& x,CI y)
{if ((x-=y)<0) x+=mod;
}
int main()
{scanf("%d",&n);int pw=n,ans=0; fact[0]=1;for (RI i=1;i<=n;++i,pw=1LL*pw*n%mod) inc(ans,pw);int m; fib[0]=0; fib[1]=1;for (RI i=2;;++i){fib[i]=fib[i-1]+fib[i-2];if (fib[i]>n) { m=i; break; }}for (RI i=1;i<=m;++i) fact[i]=1LL*fact[i-1]*i%mod;// printf("m = %d\n",m);f[1][0]=1;for (RI i=1;i<=min(n,m);++i)for (RI j=0;j<=n;++j){if (f[i][j]==0) continue;inc(f[i+1][j],f[i][j]);if (j+fib[i]<=n){inc(f[i][j+fib[i]],f[i][j]);if (j+fib[i]+fib[i-1]<=n){dec(ans,1LL*fact[i]*f[i][j]%mod);} else{dec(ans,1LL*fact[i]*f[i][j]%mod*inv2%mod);}}}return printf("%d\n",ans),0;
}
E. Enigma
经典势能分析题，考虑直接暴力跑多源最短路，每次增加一个新起点时，当且仅当它能减小其它点的答案时才松弛入队
因为第一次操作后每个点的答案 \(\le 25\)，而一个点入队会导致其答案至少减少 \(1\)，因此总的入队次数就是 \(25\times 100000\) 级别的
#include<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#define RI register int
#define CI const int&
using namespace std;
const int N=100000;
int n,dis[N+5],bkt[N+5]; vector <int> v[N+5];
int main()
{for (RI mask=0;mask<N;++mask){for (RI i=1,base=1;i<=5;++i,base*=10){int d=mask/base%10;if (d+1<=9) v[mask].push_back(mask+base);else v[mask].push_back(mask-base*9); // d == 9if (d-1>=0) v[mask].push_back(mask-base);else v[mask].push_back(mask+base*9); // d == 0}dis[mask]=N; ++bkt[dis[mask]];}// for (auto x:v[0]) printf("%d ",x); putchar('\n');for (scanf("%d",&n);n;--n){char s[10]; scanf("%s",s);int mask=0;for (RI i=0;i<5;++i)mask=mask*10+(s[i]-'0');queue <int> q; q.push(mask);auto upt=[&](int& x,CI y){--bkt[x]; x=y; ++bkt[x];};upt(dis[mask],0);while (!q.empty()){int now=q.front(); q.pop();// printf("mask = %d\n",now);for (auto to:v[now]){// printf("to = %d\n",to);if (dis[to]>dis[now]+1){upt(dis[to],dis[now]+1);q.push(to);}}}for (RI i=25;i>=0;--i)if (bkt[i]!=0){printf("%d %d\n",i,bkt[i]);break;}}return 0;
}
F. Finances
这题纯队友讨论+写出来的，我题目都没看
#include <bits/stdc++.h>using llsi = long long signed int;#define int llsiconstexpr int $n = 1'000'006;int n, m;
int a[$n];
std::vector<std::tuple<int, llsi, int>> e[$n];
std::vector<std::pair<int, llsi>> e2[$n];
int ecc[$n], ecc_cnt;
llsi ecc_a[$n];int __edge_cnt = 0;void add_edge(int u, int v, llsi c) {e[u].emplace_back(v, c, __edge_cnt++);e[v].emplace_back(u, c, __edge_cnt++);
}namespace Tarjan {int low[$n], dfn[$n];std::stack<int> st;std::vector<std::tuple<int, int, llsi>> bdg;int cnt;void tarjan(int x, std::tuple<int, llsi, int> be) {auto [fa, c, las_id] = be;low[x] = dfn[x] = ++cnt;st.push(x);for(auto [out, cc, id]: e[x]) {if(id == las_id) continue;if(!dfn[out]) {tarjan(out, {x, cc, id ^ 1});low[x] = std::min(low[x], low[out]);} else {low[x] = std::min(low[x], dfn[out]);}}if(dfn[x] == low[x]) {ecc[x] = ++ecc_cnt;ecc_a[ecc_cnt] += a[x];while(st.top() != x) {ecc[st.top()] = ecc_cnt;ecc_a[ecc_cnt] += a[st.top()];st.pop();}st.pop();if(fa != -1) bdg.emplace_back(x, fa, c);}}
}void init() {__edge_cnt = 0;Tarjan::cnt = 0;Tarjan::bdg.clear();ecc_cnt = 0;for(int i = 0; i <= n; ++i) {e[i].clear();e2[i].clear();Tarjan::low[i] = 0;Tarjan::dfn[i] = 0;ecc_a[i] = 0;}
}std::pair<bool, llsi> dfs(int cur, int fa, llsi c) {// std::cerr << std::format("dfs({} {} {})", cur, fa, c) << char(10);llsi sum = ecc_a[cur];for(auto [out, cc]: e2[cur]) if(out != fa) {auto [res_out, sum_out] = dfs(out, cur, cc);if(!res_out) return { false, 0 };sum += sum_out;}return { std::abs(sum) <= c, sum };
}bool work() {std::cin >> n >> m;init();for(int i = 1; i <= n; ++i) std::cin >> a[i];for(int i = 1, u, v; i <= m; ++i) {llsi c;std::cin >> u >> v >> c;add_edge(u, v, c);}Tarjan::tarjan(1, {-1, -1, -1});// for(int i = 1; i <= n; ++i) std::cerr << ecc[i] << char(i == n ? 10 : 32);for(auto [u, v, c]: Tarjan::bdg) {e2[ecc[u]].emplace_back(ecc[v], c);e2[ecc[v]].emplace_back(ecc[u], c);}return dfs(1, 0, 0).first;
}int32_t main() {std::ios::sync_with_stdio(false);int T; std::cin >> T; while(T--) std::cout << (work() ? "TAK\n" : "NIE\n");return 0;
}
H. Hacking
签到，我题都没看
#include <bits/stdc++.h>int main() {int n; std::cin >> n;int sum = 0;while(n--) { int a; std::cin >> a; sum += a; }std::cout << 600 - sum << char(10);return 0;
}
I. ICPC Isolation
爆搜+打表，真是公公又式式
#include <bits/stdc++.h>const char* UNAME[27] = { "UW", "UJ", "UWR", "MAP", "PW", "AGH", "PG", "NLU", "PUT", "PO", "PWR", "SGGW", "UMCS", "UR", "ZUT", "DTP", "GOO", "HUA", "KUL", "PL", "PM", "PS", "UAM", "UG", "UMK", "UO", "WAT" };
int UCOUNT[27] = { 11 , 9 , 8 , 7 , 6 , 5 , 4 , 3 , 3 , 2 , 2 , 2 , 2 , 2 , 2 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 };extern int ans[27][8][10];void print_table(const std::string &prefix, int ans[8][10]) {std::cout << " // " << prefix << char(10);for(int i = 0; i < 8; ++i) {std::cout << " ";for(int j = 0; j < 10; ++j)std::cout << std::setw(2) << ans[i][j] << ",",std::cout << char(j == 9 ? 10 : 32);}
}void print_ans() {std::cout << "int ans[27][8][10] = {\n";for(int k = 0; k < 27; ++k)print_table(std::format("ars = {} ({})", k, UNAME[k]), ans[k]);std::cout << "};\n";return ;
}bool dfs(int ans[8][10], int i, int j) {if(i == 8) return true;// std::cerr << "[dfs] i, j = " << i << ", " << j << char(10);int ni = i, nj = j + 1;if(nj == 10) nj = 0, ni += 1;for(int cl = 0; cl < 27; ++cl) if(UCOUNT[cl]) {bool res = false;;for(auto [si, sj]: (int[4][2]){{i - 1, j - 1}, {i - 1, j}, {i - 1, j + 1}, {i, j - 1}}) {if(si < 0 || si >= 8) continue;if(sj < 0 || sj >= 10) continue;if(cl == ans[si][sj]) goto __continue__;}UCOUNT[cl]--; ans[i][j] = cl;res = dfs(ans, ni, nj);UCOUNT[cl]++;if(res) return true;
__continue__: continue;}return false;
}void prep_ans() {for(int ars = 0; ars < 27; ++ars) {std::cerr << "Calculate ans for (ars = " << ars << ")" << std::endl;ans[ars][0][0] = ars;UCOUNT[ars]--;dfs(ans[ars], 0, 1);UCOUNT[ars]++;}
}int main() {std::ios::sync_with_stdio(false);// memset(ans, -1, sizeof(ans));// prep_ans();// print_ans();// return 0;std::string ars; std::cin >> ars;int k;for(k = 0; k < 27; ++k) if(ars == UNAME[k]) break;assert(k < 27);for(int i = 0; i < 8; ++i) for(int j = 0; j < 10; ++j)std::cout << UNAME[ans[k][i][j]] << char(j == 9 ? 10 : 32);return 0;
}int ans[27][8][10] = {// ars = 0 (UW)0, 1, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 2, 3, 2, 3,0, 1, 0, 1, 0, 1, 0, 1, 0, 4,2, 3, 2, 3, 2, 4, 5, 4, 5, 6,0, 4, 5, 4, 5, 6, 7, 6, 7, 4,5, 6, 7, 8, 9, 8, 9, 8, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 1 (UJ)1, 0, 1, 0, 1, 0, 1, 0, 1, 0,2, 3, 2, 3, 2, 3, 2, 3, 2, 3,0, 1, 0, 1, 0, 1, 0, 1, 0, 4,2, 3, 2, 3, 2, 4, 5, 4, 5, 6,0, 4, 5, 4, 5, 6, 7, 6, 7, 4,5, 6, 7, 8, 9, 8, 9, 8, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 2 (UWR)2, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 3, 2, 3, 2, 3, 2, 3, 2, 3,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 4, 5, 4, 5, 4,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 8, 9, 8, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 3 (MAP)3, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 4, 5, 4, 5, 4,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 8, 9, 8, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 4 (PW)4, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 8, 9, 8, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 5 (AGH)5, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,6, 7, 8, 9, 8, 4, 8, 9, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 6 (PG)6, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 7, 8, 9, 8, 4, 8, 9, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 7 (NLU)7, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 8, 9, 8, 4, 8, 9, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 8 (PUT)8, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 10, 11,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 9 (PO)9, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 10, 8, 10,11, 12, 11, 12, 13, 14, 13, 14, 15, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 10 (PWR)10, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,11, 12, 11, 12, 13, 14, 13, 14, 15, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 11 (SGGW)11, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 13, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 12 (UMCS)12, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 13, 14, 13, 14, 15, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 13 (UR)13, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 15, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 14 (ZUT)14, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 15, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 15 (DTP)15, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 16,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 16 (GOO)16, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,17, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 17 (HUA)17, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 18, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 18 (KUL)18, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 19, 20, 21, 22, 23, 24, 25, 26,// ars = 19 (PL)19, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 18, 20, 21, 22, 23, 24, 25, 26,// ars = 20 (PM)20, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 18, 19, 21, 22, 23, 24, 25, 26,// ars = 21 (PS)21, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 18, 19, 20, 22, 23, 24, 25, 26,// ars = 22 (UAM)22, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 18, 19, 20, 21, 23, 24, 25, 26,// ars = 23 (UG)23, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 18, 19, 20, 21, 22, 24, 25, 26,// ars = 24 (UMK)24, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 18, 19, 20, 21, 22, 23, 25, 26,// ars = 25 (UO)25, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 18, 19, 20, 21, 22, 23, 24, 26,// ars = 26 (WAT)26, 0, 1, 0, 1, 0, 1, 0, 1, 0,1, 2, 3, 2, 3, 2, 3, 2, 3, 2,0, 4, 0, 1, 0, 1, 0, 1, 0, 1,2, 3, 2, 3, 2, 3, 4, 5, 4, 5,0, 4, 5, 4, 5, 6, 7, 6, 7, 6,5, 6, 7, 8, 9, 4, 8, 9, 8, 10,10, 11, 12, 11, 12, 13, 14, 13, 14, 15,16, 17, 18, 19, 20, 21, 22, 23, 24, 25,
};
J. Jury of AMPPZ
考虑二分答案 \(x\)，则贪心地让计算几何题里 \((val,id)\) 二元组最大的那 \(x\) 个进入前 \(k\) 名一定最优
要最大化它们的最小值，只需要从小到大排序后依次加上 \(n,n-1,\ldots,n-x+1\) 即可
对于非计算几何题，前 \(k-x\) 大的元素其实可以忽略不管，我们只要保证上面的最小值能超过剩下题的最大值即可
还是如法炮制地贪心，从大到小排序后依次加上 \(1,2,\ldots\) 即可，最后求个最大值和前面比较一下
#include <bits/stdc++.h>using llsi = long long signed int;#define int llsiconstexpr int $n = 1'000'006;int n, m;
int a[$n];
std::vector<std::tuple<int, llsi, int>> e[$n];
std::vector<std::pair<int, llsi>> e2[$n];
int ecc[$n], ecc_cnt;
llsi ecc_a[$n];int __edge_cnt = 0;void add_edge(int u, int v, llsi c) {e[u].emplace_back(v, c, __edge_cnt++);e[v].emplace_back(u, c, __edge_cnt++);
}namespace Tarjan {int low[$n], dfn[$n];std::stack<int> st;std::vector<std::tuple<int, int, llsi>> bdg;int cnt;void tarjan(int x, std::tuple<int, llsi, int> be) {auto [fa, c, las_id] = be;low[x] = dfn[x] = ++cnt;st.push(x);for(auto [out, cc, id]: e[x]) {if(id == las_id) continue;if(!dfn[out]) {tarjan(out, {x, cc, id ^ 1});low[x] = std::min(low[x], low[out]);} else {low[x] = std::min(low[x], dfn[out]);}}if(dfn[x] == low[x]) {ecc[x] = ++ecc_cnt;ecc_a[ecc_cnt] += a[x];while(st.top() != x) {ecc[st.top()] = ecc_cnt;ecc_a[ecc_cnt] += a[st.top()];st.pop();}st.pop();if(fa != -1) bdg.emplace_back(x, fa, c);}}
}void init() {__edge_cnt = 0;Tarjan::cnt = 0;Tarjan::bdg.clear();ecc_cnt = 0;for(int i = 0; i <= n; ++i) {e[i].clear();e2[i].clear();Tarjan::low[i] = 0;Tarjan::dfn[i] = 0;ecc_a[i] = 0;}
}std::pair<bool, llsi> dfs(int cur, int fa, llsi c) {// std::cerr << std::format("dfs({} {} {})", cur, fa, c) << char(10);llsi sum = ecc_a[cur];for(auto [out, cc]: e2[cur]) if(out != fa) {auto [res_out, sum_out] = dfs(out, cur, cc);if(!res_out) return { false, 0 };sum += sum_out;}return { std::abs(sum) <= c, sum };
}bool work() {std::cin >> n >> m;init();for(int i = 1; i <= n; ++i) std::cin >> a[i];for(int i = 1, u, v; i <= m; ++i) {llsi c;std::cin >> u >> v >> c;add_edge(u, v, c);}Tarjan::tarjan(1, {-1, -1, -1});// for(int i = 1; i <= n; ++i) std::cerr << ecc[i] << char(i == n ? 10 : 32);for(auto [u, v, c]: Tarjan::bdg) {e2[ecc[u]].emplace_back(ecc[v], c);e2[ecc[v]].emplace_back(ecc[u], c);}return dfs(1, 0, 0).first;
}int32_t main() {std::ios::sync_with_stdio(false);int T; std::cin >> T; while(T--) std::cout << (work() ? "TAK\n" : "NIE\n");return 0;
}K. Kids' Blocks
又是个我没看题的签到
#include <bits/stdc++.h>constexpr int $n = 300'005;
int a[$n], pm[$n], sm[$n];int main() {std::ios::sync_with_stdio(false);int n; std::cin >> n;for(int i = 1; i <= n; ++i) std::cin >> a[i];pm[1] = a[1]; for(int i = 2; i <= n; ++i) pm[i] = std::max(pm[i - 1], a[i]);sm[n] = a[n]; for(int i = n - 1; i >= 1; --i) sm[i] = std::max(sm[i + 1], a[i]);int ans = 0x7fff'ffff;for(int i = 1; i < n; ++i) ans = std::min(ans, std::abs(sm[i + 1] - pm[i]));std::cout << ans << char(10);return 0;
}
L. Linear Averaging
我去蹲了个坑队友就会这个题了，好像是个转化成几何问题后用 convex trick 搞一搞的题，只能说队友太有实力
#include <bits/stdc++.h>using llsi = long long signed int;
using ldb = long double;#define double ldbconstexpr int $n = 500005;int n, m, k;
int a[$n];
llsi sum_k[$n];
llsi pos[$n], neg[$n], ans[$n];double ans1[$n], ans2[$n];
void solve(double ans[$n]) {std::vector<std::pair<double, double>> lines;for(llsi i = 1; i <= m; ++i) {if(pos[i] >= 0) {double K = double(pos[i]);double B = double(i);if(K == 0) {ans[i] = 0.0L;continue;}while(lines.size() && K >= lines.back().first) lines.pop_back();while(lines.size() >= 2) {auto [K1, B1] = lines[lines.size() - 1];auto [K2, B2] = lines[lines.size() - 2];auto Y1 = (K1 * B2 - K2 * B1) / (K1 - K2);auto Y2 = (K1 * B - K * B1) / (K1 - K );if(Y2 <= Y1) break;lines.pop_back(); }lines.emplace_back(K, B);}while(lines.size() >= 2) {auto [K1, B1] = lines[lines.size() - 1];auto [K2, B2] = lines[lines.size() - 2];if((K1 * B2 - K2 * B1) / (K1 - K2) >= i) break;lines.pop_back();}// std::cerr << "[DEBUG solve() i = " << i << "] lines:";// if(lines.empty()) std::cerr << " [empty]";// else for(auto [K, B]: lines) std::cerr << " (" << K << ", " << B << ")";// std::cerr << char(10);if(lines.size()) {auto [K, B] = lines.back();ans[i] = ((double)i - B) / (K / k);} else {ans[i] = 1e16;}// std::cerr << "ans[" << i << "] = " << ans[i] << char(10);}
}const char* clip(llsi a, llsi INF) {if(a >= INF) return "INF";if(a <= -INF) return "NNF";static char ss[40];std::sprintf(ss, "%lld", a);return ss;
}int main() {std::ios::sync_with_stdio(false);std::cerr << std::fixed << std::setprecision(2);std::cin >> n >> m >> k;for(int i = 1; i <= n; ++i) std::cin >> a[i];/* prep sum_k */ {llsi s = 0;for(int i = 1; i < k; ++i) s += a[i];for(int i = k; i <= n; ++i) {s += a[i];sum_k[i - k + 1] = s;s -= a[i - k + 1];}}std::deque<int> q_pos, q_neg;for(int i = 1; i <= m; ++i)pos[i] = -4e18,neg[i] = 4e18;for(int i = 1; i <= n; ++i)pos[a[i]] = llsi(k) * llsi(a[i]),neg[a[i]] = llsi(k) * llsi(a[i]);for(int i = 1; i <= n; ++i) {if(i + k - 1 <= n) {while(q_pos.size() && sum_k[i] >= sum_k[q_pos.back()]) q_pos.pop_back();while(q_neg.size() && sum_k[i] <= sum_k[q_neg.back()]) q_neg.pop_back();q_pos.push_back(i);q_neg.push_back(i);}while(q_pos.size() && q_pos.front() + k - 1 < i) q_pos.pop_front();while(q_neg.size() && q_neg.front() + k - 1 < i) q_neg.pop_front();pos[a[i]] = std::max(pos[a[i]], sum_k[q_pos.front()]);neg[a[i]] = std::min(neg[a[i]], sum_k[q_neg.front()]);}for(llsi i = 1; i <= m; ++i)pos[i] -= i * k,neg[i] -= i * k; solve(ans1);for(int i = 1; i <= m; ++i) pos[i] = -neg[m - i + 1];solve(ans2);// std::cerr << "[DEBUG] pos[]: "; for(int i = 1; i <= m; ++i) std::cerr << clip(pos[i], 100000) << char(i == m ? 10 : 32);// std::cerr << "[DEBUG] neg[]: "; for(int i = 1; i <= m; ++i) std::cerr << clip(neg[i], 100000) << char(i == m ? 10 : 32);// std::cerr << "[DEBUG] ans1[]: "; for(int i = 1; i <= m; ++i) std::cerr << ans1[i] << char(i == m ? 10 : 32);// std::cerr << "[DEBUG] ans2[]: "; for(int i = 1; i <= m; ++i) std::cerr << ans2[i] << char(i == m ? 10 : 32);std::cout << std::fixed << std::setprecision(20);for(int i = 1; i <= m; ++i) {double ans = std::min(ans1[i], ans2[m - i + 1]);if(ans < 0.0L || ans > 1.0L) std::cout << "-1\n";else std::cout << ans << char(10);}return 0;
}
Postscript
感觉两天一训强度还是挺大的，不禁让我想起之前每天一训是怎么坚持下来的
】详解Linux文本编辑器：Vim从入门到精通——完整教程与实战指南（上） - 详解)
