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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; do{c=getchar();if ('-'==c)f=-1;}while(!isdigit(c)&&c!=EOF); if(c==EOF)return false;do{ans=(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;do{rnext();if('-'==*ps)f=-1;} while(!isdigit(*ps)&&ps!=pe);if(ps==pe)return false;do{ans=(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; do{c=getchar();if ('-'==c)f=-1;}while(!isdigit(c)&&c!=EOF); if(c==EOF)return false;do{ans=(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;do{rnext();if('-'==*ps)f=-1;} while(!isdigit(*ps)&&ps!=pe);if(ps==pe)return false;do{ans=(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; }