2019 CCPC网络赛
Posted hugh-locke
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一到网络赛，大家都是东亚之光
1001
00:23:46 solved by hl
签到 枚举一下位就行了
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read()int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘)if (c == ‘-‘) f = -1;c = getchar(); while (c >= ‘0‘&&c <= ‘9‘)x = x * 10 + c - ‘0‘;c = getchar();return x*f; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; LL A,B; int main() int T = read(); while(T--) scanf("%lld%lld",&A,&B); LL c = 0; for(int i = 0; i <= 32; i ++) LL t = (1ll << i); int a = 0,b = 0; if(A & t) a = 1; if(B & t) b = 1; if(a && b) c |= t; if(!c) c++; Prl(c); return 0;
1002
2:40:10(-1) solved by hl,ysw
对序列维护一个权值线段树，维护的东西是区间最大值，然后对于每一个查询二分查找答案，如果在起始点在答案有东西的下标大于终点，就代表答案可以更小
时间复杂度n(logn)²,事实上由于权值线段树的特性，寻找大于R的最小值可以直接在线段树上操作，简化了二分这个操作，时间复杂度nlogn
比赛的时候并没有想到，冲了读入挂趁评测机不注意过去了
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; const int MAXBUF=10000;char buf[MAXBUF],*ps=buf,*pe=buf+1; inline bool isdigit(const char& n) return (n>=‘0‘&&n<=‘9‘); inline void rnext()if(++ps==pe)pe=(ps=buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin); template <class T> inline bool in(T &ans) #ifdef VSCode ans=0;T f=1;register char c; doc=getchar();if (‘-‘==c)f=-1;while(!isdigit(c)&&c!=EOF); if(c==EOF)return false;doans=(ans<<1)+(ans<<3)+c-48; c=getchar();while(isdigit(c)&&c!=EOF);ans*=f;return true; #endif #ifndef VSCode ans =0;T f=1;if(ps==pe)return false;dornext();if(‘-‘==*ps)f=-1; while(!isdigit(*ps)&&ps!=pe);if(ps==pe)return false;doans=(ans<<1)+(ans<<3)+*ps-48; rnext();while(isdigit(*ps)&&ps!=pe);ans*=f;return true; #endif const int MAXOUT=10000; char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT; inline void write()fwrite(bufout,sizeof(char),pout-bufout,stdout);pout = bufout; inline void out_char(char c)*(pout++)=c;if(pout==pend)write(); inline void out_str(char *s)while(*s)*(pout++)=*(s++);if(pout==pend)write(); template <class T>inline void out_int(T x) if(!x)out_char(‘0‘);return; if(x<0)x=-x,out_char(‘-‘);int len=0;while(x)outtmp[len++]=x%10+48;x/=10;outtmp[len]=0; for(int i=0,j=len-1;i<j;i++,j--) swap(outtmp[i],outtmp[j]);out_str(outtmp); template<typename T, typename... T2> inline int in(T& value, T2&... value2) in(value); return in(value2...); int read()int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘)if (c == ‘-‘) f = -1;c = getchar(); while (c >= ‘0‘&&c <= ‘9‘)x = x * 10 + c - ‘0‘;c = getchar();return x*f; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int a[maxn],id[maxn]; struct Tree int l,r,Max; tree[maxn << 2]; inline void Pushup(int t) tree[t].Max = max(tree[t << 1].Max,tree[t << 1 | 1].Max); inline void Build(int t,int l,int r) tree[t].l = l; tree[t].r = r; if(l == r) tree[t].Max = id[l]; return; int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); Pushup(t); inline int query(int t,int l,int r) if(l <= tree[t].l && tree[t].r <= r) return tree[t].Max; int m = (tree[t].l + tree[t].r) >> 1; if(r <= m) return query(t << 1,l,r); else if(l > m) return query(t << 1 | 1,l,r); return max(query(t << 1,l,m),query(t << 1 | 1,m + 1,r)); inline void update(int t,int p) if(tree[t].l == tree[t].r) tree[t].Max = INF; return; int m = (tree[t].l + tree[t].r) >> 1; if(p <= m) update(t << 1,p); else update(t << 1 | 1,p); Pushup(t); inline int solve(int k,int R) int l = k,r = N + 1; int ans = N + 1; while(l <= r) int m = l + r >> 1; if(query(1,k,m) > R) ans = m; r = m - 1; else l = m + 1; return ans; int main() int T; in(T); while(T--) in(N); in(M); for(int i = 1; i <= N ; i ++) in(a[i]),id[a[i]] = i; id[N + 1] = INF; Build(1,1,N + 1); int ans = 0; while(M--) int op; in(op); if(op == 1) int t; in(t); t ^= ans; update(1,a[t]); else int r,k; in(r); in(k); r ^= ans; k ^= ans; // cout << k << " " << r << endl; Pri(ans = solve(k,r)); return 0;
1003
1:31:38 (-1) solved by hl
主席树的tot没有初始化 RE一发
先走一个后缀数组，发现只要在给定起始位置上的Height数组连续长度超过给定子串长度len中的一段
用ST表 + 二分可以log时间求出这一段的左右端点，然后就是一个静态区间K大，主席树即可
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read()int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘)if (c == ‘-‘) f = -1;c = getchar(); while (c >= ‘0‘&&c <= ‘9‘)x = x * 10 + c - ‘0‘;c = getchar();return x*f; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; char str[maxn]; int sa[maxn],rak[maxn],tex[maxn],tp[maxn],Height[maxn]; void GetHeight() int j, k = 0; for(int i = 1; i <= N; i++) if(k) k--; int j = sa[rak[i] - 1]; while(str[i + k] == str[j + k]) k++; Height[rak[i]] = k; void Qsort() for(int i = 0; i <= M ; i ++) tex[i] = 0; for(int i = 1; i <= N ; i ++) tex[rak[i]]++; for(int i = 1; i <= M ; i ++) tex[i] += tex[i - 1]; for(int i = N; i >= 1 ; i --) sa[tex[rak[tp[i]]]--] = tp[i]; void SA() for(int i = 1; i <= N ; i ++) rak[i] = str[i] - ‘a‘ + 1,tp[i] = i; Qsort(); for(int w = 1,p = 0; p < N; w <<= 1,M = p) p = 0; for(int i = 1; i <= w; i ++) tp[++p] = N - w + i; for(int i = 1; i <= N ; i ++) if(sa[i] > w) tp[++p] = sa[i] - w; Qsort(); swap(tp,rak); rak[sa[1]] = p = 1; for(int i = 2; i <= N ; i ++) rak[sa[i]] = (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])?p:++p; int T[maxn]; struct Tree int lt,rt,sum; void init() lt = rt = sum = 0; tree[maxn * 32]; int tot; void newnode(int &t) t = ++tot; tree[t].init(); void Build(int &t,int l,int r) newnode(t); if(l == r) return; int m = l + r >> 1; Build(tree[t].lt,l,m); Build(tree[t].rt,m + 1,r); void update(int &t,int pre,int l,int r,int p) newnode(t); tree[t] = tree[pre]; tree[t].sum++; if(l == r) return; int m = l + r >> 1; if(p <= m) update(tree[t].lt,tree[pre].lt,l,m,p); else update(tree[t].rt,tree[pre].rt,m + 1,r,p); int query(int L,int R,int l,int r,int k) if(k > tree[R].sum - tree[L].sum) return -1; if(l >= r) return l; int num = tree[tree[R].lt].sum - tree[tree[L].lt].sum; int m = l + r >> 1; if(num >= k) return query(tree[L].lt,tree[R].lt,l,m,k); else return query(tree[L].rt,tree[R].rt,m + 1,r,k - num); const int SP = 20; int MIN[maxn][SP]; int mm[maxn]; void initRMQ(int n,int b[]) for(int i = 1; i <= n ; i ++) for(int j = 0; j < SP; j ++) MIN[i][j] = INF; mm[0] = -1; for(int i = 1; i <= n ; i ++) mm[i] = ((i & (i - 1)) == 0)?mm[i - 1] + 1:mm[i - 1]; MIN[i][0] = b[i]; for(int j = 1; j <= mm[n]; j ++) for(int i = 1; i + (1 << j) - 1 <= n ; i ++) MIN[i][j] = min(MIN[i][j - 1],MIN[i + (1 << (j - 1))][j - 1]); int rmq(int x,int y) if(x > y) return -1; int k = mm[y - x + 1]; return min(MIN[x][k],MIN[y - (1 << k) + 1][k]); int main() int TTT = read(); while(TTT--) Sca2(N,K); tot = 0; scanf("%s",str + 1); M = 122; SA(); GetHeight(); Build(T[0],1,N); for(int i = 1; i <= N ; i ++) update(T[i],T[i - 1],1,N,sa[i]); initRMQ(N,Height); while(K--) int l,r,k; Sca3(l,r,k); int len = r - l + 1; int p = rak[l]; int L = p + 1,R = p; int ll = 2,rr = p; while(ll <= rr) int mid = ll + rr >> 1; if(rmq(mid,p) >= len) rr = mid - 1; L = mid; else ll = mid + 1; L--; ll = p + 1,rr = N; while(ll <= rr) int mid = ll + rr >> 1; if(rmq(p + 1,mid) >= len) ll = mid + 1; R = mid; else rr = mid - 1; Pri(query(T[L - 1],T[R],1,N,k)); return 0;
1004
3:37:37(-2) solved by hl
做法
1.离线所有询问
2.将一个(点，当前距离)的二元组不断扔进堆，每次取出一个二元组然后丢进所有这个点引申出去的二元组
3.直到取出前50000个，所有答案就都出来了
注意点
1.堆内只能容忍最多5w个二元组，不限制会MLE，所以要用multiset弹出最大的
2.菊花图可以轻松卡掉，如果不边排序的话
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; const int MAXBUF=10000;char buf[MAXBUF],*ps=buf,*pe=buf+1; inline bool isdigit(const char& n) return (n>=‘0‘&&n<=‘9‘); inline void rnext()if(++ps==pe)pe=(ps=buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin); template <class T> inline bool in(T &ans) #ifdef VSCode ans=0;T f=1;register char c; doc=getchar();if (‘-‘==c)f=-1;while(!isdigit(c)&&c!=EOF); if(c==EOF)return false;doans=(ans<<1)+(ans<<3)+c-48; c=getchar();while(isdigit(c)&&c!=EOF);ans*=f;return true; #endif #ifndef VSCode ans =0;T f=1;if(ps==pe)return false;dornext();if(‘-‘==*ps)f=-1; while(!isdigit(*ps)&&ps!=pe);if(ps==pe)return false;doans=(ans<<1)+(ans<<3)+*ps-48; rnext();while(isdigit(*ps)&&ps!=pe);ans*=f;return true; #endif const int MAXOUT=10000; char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT; inline void write()fwrite(bufout,sizeof(char),pout-bufout,stdout);pout = bufout; inline void out_char(char c)*(pout++)=c;if(pout==pend)write(); inline void out_str(char *s)while(*s)*(pout++)=*(s++);if(pout==pend)write(); template <class T>inline void out_int(T x) if(!x)out_char(‘0‘);return; if(x<0)x=-x,out_char(‘-‘);int len=0;while(x)outtmp[len++]=x%10+48;x/=10;outtmp[len]=0; for(int i=0,j=len-1;i<j;i++,j--) swap(outtmp[i],outtmp[j]);out_str(outtmp); template<typename T, typename... T2> inline int in(T& value, T2&... value2) in(value); return in(value2...); #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 5e4 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,Q; struct Edge int to; LL dis; Edge() Edge(int to,LL dis):to(to),dis(dis) ; bool cmp(Edge a,Edge b) return a.dis < b.dis; vector<Edge>MAP[maxn]; PII query[maxn]; struct node int pos; LL dis; node() node(int pos,LL dis):pos(pos),dis(dis) friend bool operator < (node a,node b) return a.dis < b.dis; ; LL ans[maxn]; int main() int T; in(T); while(T--) in(N); in(M); in(Q); for(int i = 0 ; i <= N; i ++) MAP[i].clear(); for(int i = 1; i <= M ; i ++) int u,v; LL w; in(u); in(v); in(w); MAP[u].push_back(Edge(v,w)); for(int i = 1; i <= N ; i ++) sort(MAP[i].begin(),MAP[i].end(),cmp); for(int i = 1; i <= Q; i ++) in(query[i].fi); query[i].se = i; sort(query + 1,query + 1 + Q); multiset<node>P; for(int i = 1; i <= N ; i ++) for(int j = 0 ; j < MAP[i].size(); j ++) P.insert(node(MAP[i][j].to,MAP[i][j].dis)); // cout << MAP[i][j].to << " " << MAP[i][j].dis << endl; multiset<node>::iterator it; int sum = query[Q].fi; LL ed = 4e18; while(P.size() > sum) it = P.end(); it--;ed = (*it).dis; P.erase(it); int k = 0; int cnt = 1; while(sum--) it = P.begin(); node u = *it; k++; P.erase(it); while(cnt <= Q && query[cnt].fi == k) ans[query[cnt].se] = u.dis,cnt++; for(int i = 0 ; i < MAP[u.pos].size(); i ++) int v = MAP[u.pos][i].to; if(MAP[u.pos][i].dis + u.dis >= ed) break; P.insert(node(v,MAP[u.pos][i].dis + u.dis)); while(P.size() > sum) it = P.end(); it--; ed = (*it).dis; P.erase(it); for(int i = 1; i <= Q; i ++) Prl(ans[i]); return 0;
1006
0:18:13 solved by hl
签到，从后往前依次输出即可
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read()int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘)if (c == ‘-‘) f = -1;c = getchar(); while (c >= ‘0‘&&c <= ‘9‘)x = x * 10 + c - ‘0‘;c = getchar();return x*f; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; bool vis[maxn]; int a[maxn],b[maxn]; int main() while(~Sca2(N,M)) for(int i = 0; i <= N ; i ++) vis[i] = 0; for(int i = 1; i <= N ; i ++) Sca(a[i]); for(int i = 1; i <= M ; i ++) Sca(b[i]); for(int i = M ; i >= 1; i --) if(!vis[b[i]]) printf("%d ",b[i]); vis[b[i]] = 1; for(int i = 1; i <= N ; i ++) if(!vis[a[i]]) printf("%d ",a[i]); vis[a[i]] = 1; return 0;
1007
0:13:46 solved by hl
签到，按照规律弄出一个大的矩阵然后输出即可
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read()int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘)if (c == ‘-‘) f = -1;c = getchar(); while (c >= ‘0‘&&c <= ‘9‘)x = x * 10 + c - ‘0‘;c = getchar();return x*f; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 2010; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int MAP[maxn][maxn]; int main() MAP[1][1] = MAP[1][2] = MAP[2][2] = 1; MAP[2][1] = 0; for(int i = 2; i <= 10; i ++) int s = (1LL << (i - 1)); int t = s; for(int j = 1; j <= s; j ++) for(int k = 1; k <= s; k ++) MAP[j][k + s] = MAP[j][k]; MAP[j + s][k + s] = MAP[j][k]; MAP[j + s][k] = MAP[j][k] ^ 1; int T = read(); while(T--) Sca(N); int s = (1LL << N); for(int i = 1; i <= s; i ++) for(int j = 1; j <= s; j ++) if(MAP[i][j]) printf("C"); else putchar(‘P‘); puts(""); return 0;
1008
1:54:25(-1) solved by zcz
因hl冲了一发贪心而产生的WA
队友写的 我不太明白怎么做
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; long long a[100005]; long long b[100005]; int main() int T; cin>>T; while(T--) int n,k; cin>>n>>k; long long s1=0; long long s2=0; for(int i=0;i<n;i++) scanf("%lld",&a[i]); s1+=a[i]/k; s2+=a[i]; b[i]=k-(a[i]%k); sort(b,b+n); for(int i=0;i<n-(s1+1);i++) s2+=b[i]; cout<<s2+k<<endl; return 0;
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