上海理工大学联想杯 4题滚粗记
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果然是被生活压垮的人，即将退役和告别大学生活了
早上放下了GRE来看题，写了两个小时就又不得不去学gre了，希望330+
比赛开始20分钟之后才到自习室，然后这个时候排名已经1100开外了
第一题，一看，问最长的等差数列长度，给了公差，所以拿map记录一下就好了
#include <bits/stdc++.h> using namespace std; #define limit (2000000 + 5)//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0); #define ff(a) printf("%d\\n",a ); #define pi(a,b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\\\Users\\\\tiany\\\\CLionProjects\\\\akioi\\\\data.txt", "rt", stdin) #define FOUT freopen("C:\\\\Users\\\\tiany\\\\CLionProjects\\\\akioi\\\\dabiao.txt", "wt", stdout) #define debug(x) cout<<x<<endl typedef long long ll; typedef unsigned long long ull; char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf; inline ll read(){ #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0;char s = getchar(); while(s > \'9\' || s < \'0\' ){if(s == \'-\')sign = -1;s = getchar();} while(s >= \'0\' && s <= \'9\'){x = (x << 3) + (x << 1) + s - \'0\';s = getchar();} return x * sign; #undef getchar }//快读 void print(ll x) { if(x/ 10) print(x / 10); *O++=x % 10+\'0\'; } void write(ll x, char c = \'t\') { if(x < 0)putchar(\'-\'),x = -x; print(x); if(!isalpha(c))*O++ = c; fwrite(obuf,O-obuf,1,stdout); O = obuf; } int kase; int n, m,k; int a[limit]; void solve(){ cin>>n>>k; map<int, int>mp; rep(i,1,n){ cin>>a[i]; } sort(a + 1, a + 1 + n); ll ans = 1; rep(i,1,n){ if(!mp.count(a[i])){ mp[a[i]] = mp[a[i] - k] + 1; ans = max(ans, 1ll * mp[a[i]]); } } cout<<ans<<endl; } int32_t main() { #ifdef LOCAL FOPEN; //FOUT; #endif //FASTIO //kase = read(); //while (kase--) solve(); cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\\n"; return 0; }
然后看了下K，问最大匹配，数据还小得一批，笑了，暴力加边，dinic水了过去
#include <bits/stdc++.h> using namespace std; #define limit (200000 + 5)//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0); #define ff(a) printf("%d\\n",a ); #define pi(a,b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\\\Users\\\\tiany\\\\CLionProjects\\\\akioi\\\\data.txt", "rt", stdin) #define FOUT freopen("C:\\\\Users\\\\tiany\\\\CLionProjects\\\\akioi\\\\dabiao.txt", "wt", stdout) #define debug(x) cout<<x<<endl typedef long long ll; typedef unsigned long long ull; char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf; inline ll read(){ #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0;char s = getchar(); while(s > \'9\' || s < \'0\' ){if(s == \'-\')sign = -1;s = getchar();} while(s >= \'0\' && s <= \'9\'){x = (x << 3) + (x << 1) + s - \'0\';s = getchar();} return x * sign; #undef getchar }//快读 void print(ll x) { if(x/ 10) print(x / 10); *O++=x % 10+\'0\'; } void write(ll x, char c = \'t\') { if(x < 0)putchar(\'-\'),x = -x; print(x); if(!isalpha(c))*O++ = c; fwrite(obuf,O-obuf,1,stdout); O = obuf; } int kase; int n,m,vs,ve,p; int layer[limit],head[limit], cnt; struct node{ int to ,next; ll flow, w; }edge[limit]; ll max_flow; void add_one(int u , int v, ll flow = 0){ edge[cnt].to = v; edge[cnt].next = head[u]; edge[cnt].flow = flow; edge[cnt].w = 0; head[u] = cnt++; } inline void add(int u, int v, ll flow){ add_one(u,v,flow); add_one(v, u,0); } inline void init(bool flag = true){ if(flag){ memset(head, -1, sizeof(head)); cnt = 0; }else{ memset(layer, -1, sizeof(layer)); } } inline bool bfs(){ init(false); queue<int>q; layer[vs] = 0;//从第0层开始 q.push(vs); while (q.size()){ int u = q.front(); q.pop(); traverse(u){ int v = edge[i].to,flow = edge[i].flow; if(layer[v] == -1 && flow > 0){ layer[v] = layer[u] + 1;//迭代加深 q.push(v); } } } return ~layer[ve]; } ll dfs(int u, ll flow){ if(u == ve)return flow; ll rev_flow = 0,min_flow; traverse(u){ int v =edge[i].to; ll t_flow = edge[i].flow; if(layer[v] == layer[u] + 1 && t_flow > 0){ min_flow = dfs(v, min(flow, t_flow)); flow -= min_flow; edge[i].flow -= min_flow; rev_flow += min_flow; edge[i^1].flow += min_flow; if(!flow)break; } } if(!rev_flow)layer[u] = -1; return rev_flow; } void dinic(){ while (bfs()){ max_flow += dfs(vs,inf); } } int a[limit], b[limit]; void solve(){ init(); n = read(); vs = 90001, ve = vs + 1; rep(i,1,n){ a[i] = read(); add(vs, i, 1); } rep(i,1,n){ b[i] = read(); add(i + n, ve, 1); } rep(i,1,n){ rep(j,1,n){ if(__gcd(a[i] , b[j]) != 1){ add(i, j + n, INF); } } } dinic(); write(max_flow); } int32_t main() { #ifdef LOCAL FOPEN; //FOUT; #endif //FASTIO //kase = read(); //while (kase--) solve(); cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\\n"; return 0; }
然后C题，问匹配，好像列出来之后只有6种情况能对答案产生贡献，那么就手动模拟一下就好了
#include <bits/stdc++.h> using namespace std; #define limit (2000000 + 5)//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0); #define ff(a) printf("%d\\n",a ); #define pi(a,b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\\\Users\\\\tiany\\\\CLionProjects\\\\akioi\\\\data.txt", "rt", stdin) #define FOUT freopen("C:\\\\Users\\\\tiany\\\\CLionProjects\\\\akioi\\\\dabiao.txt", "wt", stdout) #define debug(x) cout<<x<<endl typedef long long ll; typedef unsigned long long ull; char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf; inline ll read(){ #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0;char s = getchar(); while(s > \'9\' || s < \'0\' ){if(s == \'-\')sign = -1;s = getchar();} while(s >= \'0\' && s <= \'9\'){x = (x << 3) + (x << 1) + s - \'0\';s = getchar();} return x * sign; #undef getchar }//快读 void print(ll x) { if(x/ 10) print(x / 10); *O++=x % 10+\'0\'; } void write(ll x, char c = \'t\') { if(x < 0)putchar(\'-\'),x = -x; print(x); if(!isalpha(c))*O++ = c; fwrite(obuf,O-obuf,1,stdout); O = obuf; } int kase; int n,m,k; int ca,cat,c,at,t,a; void solve(){ cin>>n; rep(i,1,n) { string str; cin >> str; if (str == "ca")++ca; if (str == "cat")++cat; if (str == "a")++a; if (str == "at")++at; if (str == "t")++t; if (str == "c")++c; } // ca + t ll ans = min(ca, t), tmp; t -= ans, ca -= ans; // c + at tmp = min(c, at); ans += tmp; c -= tmp, at -= tmp; // cat ans += cat, cat = 0; // c + a + t tmp = min({c , a , t}); ans += tmp; c -= tmp, a -= tmp, t -= tmp; cout<<ans<<endl; } int32_t main() { #ifdef LOCAL FOPEN; //FOUT; #endif FASTIO //kase = read(); //while (kase--) solve(); cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\\n"; return 0; }
然后看了看B，发现题意太过复杂，虽然感觉不难，但是念及自己可怜的分数还是算了，开了J
然后发现J的范围是1e7，很典型的素数筛题，想到根号分解数字，找到最小质因数，然后log计算贡献，再从增量里把这些剪掉就行，暴力加答案。
然后好家伙，tle了，我跟队友说不要写了，这题有坑，然后队友过了一会儿告诉我，tle个锤子，他过了。
然后我发现，好像__int128没关上，那没事了，果断提交并且AC，话说我这个__int128为了谨慎起见还开了fread，没想到这么毒瘤
#include <bits/stdc++.h> using namespace std; #define limit (20000000 + 5)//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0); #define ff(a) printf("%d\\n",a ); #define pi(a,b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\\\Users\\\\tiany\\\\CLionProjects\\\\akioi\\\\data.txt", "rt", stdin) #define FOUT freopen("C:\\\\Users\\\\tiany\\\\CLionProjects\\\\akioi\\\\dabiao.txt", "wt", stdout) #define debug(x) cout<<x<<endl typedef long long ll; typedef unsigned long long ull; char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf; inline ll read(){ #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0;char s = getchar(); while(s > \'9\' || s < \'0\' ){if(s == \'-\')sign = -1;s = getchar();} while(s >= \'0\' && s <= \'9\'){x = (x << 3) + (x << 1) + s - \'0\';s = getchar();} return x * sign; #undef getchar }//快读 void print(ll x) { if(x/ 10) print(x / 10); *O++=x % 10+\'0\'; } void write(ll x, char c = \'t\') { if(x < 0)putchar(\'-\'),x = -x; print(x); if(!isalpha(c))*O++ = c; fwrite(obuf,O-obuf,1,stdout); O = obuf; } int kase; ll n,m, k; ll prime[limit], num[limit],tot; int a[limit], b[limit]; void get_prime( const int &N=2e7){ memset(num, 1, sizeof(num)); tot = 0; rep(i ,2,N){ if(num[i])prime[++tot] = i; for(int j = 1 ; j <= tot && i * prime[j] <= N ; ++j){ num[i * prime[j]] = 0; if(i % prime[j] == 0)break;//线性筛 } } } void process(ll x){ if(num[x]){ a[x] = x; b[x] = 1; return; } for(ll i = 2 ; i * i <= x; ++i){ if(num[i] and x % i == 0){ a[x] = i; ll tmp = x; while(tmp % i == 0){ b[x]++; tmp /= i; } return; } } } void solve(){ get_prime(); n = read(),m = read(), k= read(); rep(i,1 + k,n + k){ process(i); } ll ans = 0; rep(i,1 + k,n + k){ ll ti = b[i] - (b[i] / m); ll tmp = i; rep(j,1,ti){ tmp /= a[i]; } ans += i - tmp; } write(ans); } int32_t main() { #ifdef LOCAL FOPEN; //FOUT; #endif FASTIO //kase = read(); //while (kase--) solve(); cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\\n"; return 0; }
然后看了看I题，好像三分写假了，那就撤退好了，今天还有新的任务呢。
感觉之后还是多一些训练，希望有生之年能够成全我一次吧，今年ICPC加油！
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