2019 ICPC 南京网络赛
A. 00:40:35 solved by hl
很显然需要一种方法O(1)定位x,y上的数字，利用矩阵的规律可以找到
然后就是一个子矩阵和问题，小范围可以直接二维前缀和，大范围就是一个二位偏序问题，
将一个询问拆成4个询问，树状数组解决即可
#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 = 1e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,Q; LL pre[maxn]; void init(){ pre[0] = 0; for(int sum = N,i = 1; sum > 0; i ++,sum -= 2){ pre[i] = pre[i - 1] + sum + sum + sum + sum - 4; } } inline int min(int a,int b,int c,int d){ return min(min(min(a,b),c),d); } inline LL id(int x,int y){ int q = min(x,y,N - x + 1,N - y + 1); LL ans = pre[q - 1] + 1; int sx = N - q + 1,sy = N - q + 1; if(sx == x) return ans + sy - y; ans += sy - q; sy = q; if(sy == y) return ans + sx - x; ans += sx - q; sx = q; if(sx == x) return ans + y - sy; ans += (N - q + 1) - sy; sy = N - q + 1; return ans + x - sx; } inline LL val(int x,int y){ LL sum = id(x,y); LL ans = 0; while(sum){ ans += sum % 10; sum /= 10; } return ans; } struct Good{ int x,y; LL ans; }g[maxn]; struct Query{ int x,y,id,flag; Query(int x = 0,int y = 0,int id = 0,int flag = 0):x(x),y(y),id(id),flag(flag){} }query[maxn]; LL fans[maxn]; bool cmp(Good a,Good b){ return a.x < b.x; } LL tree[maxn]; void add(int u,int v){ for(;u <= N; u += u & -u) tree[u] += v; } LL getsum(int u){ LL ans = 0; for(;u > 0; u -= u & -u) ans += tree[u]; return ans; } bool cmp2(Query a,Query b){ return a.x < b.x; } int main(){ int T = read(); while(T--){ Mem(tree,0); Sca3(N,M,Q);init(); for(int i = 1; i <= Q; i ++) fans[i] = 0; for(int i = 1; i <= M; i ++){ g[i].x = read(),g[i].y = read(); g[i].ans = val(g[i].x,g[i].y); } int cnt = 0; for(int i = 1; i <= Q; i ++){ int x1 = read(),y1 = read(),x2 = read(),y2 = read(); query[++cnt] = Query(x1 - 1,y1 - 1,i,1); query[++cnt] = Query(x2,y1 - 1,i,-1); query[++cnt] = Query(x1 - 1,y2,i,-1); query[++cnt] = Query(x2,y2,i,1); } sort(g + 1,g + 1 + M,cmp); sort(query + 1,query + 1 + cnt,cmp2); int s1 = 1,s2 = 1; for(int i = 0; i <= N ; i ++){ while(s1 <= M && g[s1].x == i){ add(g[s1].y,g[s1].ans); s1++; } while(s2 <= cnt && query[s2].x == i){ fans[query[s2].id] += query[s2].flag * getsum(query[s2].y); s2++; } } for(int i = 1; i <= Q; i ++) Prl(fans[i]); } return 0; }
B.1:42:29(-1) solved by gbs
队友说 是裸的指数循环节
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef long long LL; int a,b,m; LL eular(LL n) { LL ans = n; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { ans -= ans / i; while (n % i == 0) n /= i; } } if (n != 1) ans -=ans / n; return ans; } const LL mod = 1e9 + 7; //¿ìËÙÃÝ int quick_mi(LL a, LL k,LL mod) { if (k == 0) return 1; if (a<=1) return a; bool if_over = false; LL k1 = 1; for (int i=0; i<k; i++) { k1 = k1*a; if (k1>= mod) { if_over = true; break; } } LL ans = 1; while (k) { if (k & 1) { ans *= a; ans %= mod; } a *= a; a %= mod; k >>= 1; } if (if_over) { ans += mod; } return ans; } LL dfs(int a,int b,LL mod) { if (b == 0) return 1; if (b == 1) return a; if (mod == 1) return 1; LL h1=quick_mi(a,dfs(a,b-1,eular(mod)),mod); //cout<<mod<<' '<<h1<<endl; return h1; } int main() { int t; cin >>t; while(t--) { cin >>a>>b>>m; int h1 = dfs(a,b,m); cout<<h1%m<<endl;; } #ifdef VSCode system("pause"); #endif return 0; }
D.00:44:59 solved by zcz
#include<iostream> #include<cstring> #include<cstdio> #include<vector> using namespace std; double p1[100005],p2[100005]; int n,m; vector<int> e[100005]; double getp2(int u) { if(p2[u]>=0) return p2[u]; int l=e[u].size(); double rec=0; for(int i=0;i<l;i++) { rec+=getp2(e[u][i])+1; } p2[u]=(rec+1)/l; return p2[u]; } double getp1(int u) { if(p1[u]>=0) return p1[u]; int l=e[u].size(); double rec=0; for(int i=0;i<l;i++) { rec+=getp1(e[u][i])+p2[e[u][i]]; } p1[u]=(rec+p2[u])/l; return p1[u]; } int main() { int T; cin>>T; while(T--) { cin>>n>>m; for(int i=1;i<=n;i++) { p1[i]=-1,p2[i]=-1; e[i].clear(); } p2[n]=1; p1[n]=0; while(m--) { int u,v; scanf("%d%d",&u,&v); e[u].push_back(v); } getp2(1); getp1(1); printf("%.2lf\n",p1[1]); } return 0; }
F. 1:30:28 solved by zcz and hl
线段树维护区间最大值的下标，按照数的大小从小到大询问然后修改
#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; int a[maxn],id[maxn]; int ans[maxn]; struct node{ int pos,val; node(int pos = 0,int val = 0):pos(pos),val(val){} friend node operator + (node a,node b){ if(a.val < b.val) return b; return a; } }; struct Tree{ int l,r; node ans; }tree[maxn << 2]; void Pushup(int t){ tree[t].ans = tree[t << 1].ans + tree[t << 1 | 1].ans; } void Build(int t,int l,int r){ tree[t].l = l ; tree[t].r = r; tree[t].ans = node(0,0); if(l == r){ tree[t].ans = node(l,0); return; } int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); Pushup(t); } node query(int t,int l,int r){ if(l <= tree[t].l && tree[t].r <= r){ return tree[t].ans; } 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); else{ return query(t << 1,l,m) + query(t << 1 | 1,m + 1,r); } } void update(int t,int pos){ if(tree[t].l == tree[t].r){ tree[t].ans.val = a[pos]; return; } int m = (tree[t].l + tree[t].r) >> 1; if(pos <= m) update(t << 1,pos); else update(t << 1 | 1,pos); Pushup(t); } int main(){ int T = read(); while(T--){ Sca2(N,K); for(int i = 1; i <= N; i ++) ans[i] = 0; for(int i = 1; i <= N ; i ++) Sca(a[i]),id[a[i]] = i; Build(1,1,N); for(int i = 1; i <= N ; i ++){ int l = max(1,id[i] - K),r = min(N,id[i] + K); node t = query(1,l,r); ans[i] = ans[t.val] + 1; update(1,id[i]); } for(int i = 1; i <= N; i ++){ printf("%d",ans[i]); if(i != N) printf(" "); } puts(""); } return 0; }
G.4:35:30(-7) solved by zcz
一个很麻烦的线性规划
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; long long f(long long a,long long b,long long c) { if(a<b) { long long t=a; a=b; b=t; } if(c<0) return 0; if(c>=a+b) return (a+1)*(b+1); if(c<=b) return (c+1)*(c+2)/2; if(b<c&&c<a) return (b+1)*(b+2)/2+(c-b)*(b+1); long long t=a+b-c; return (a+1)*(b+1)- t*(t+1)/2; } long long f2(long long x) { return (3*x*(x+1)+x*(x+1)*(2*x+1))/12; } long long f3(long long x) { return x*(x+1)/2; } long long f4(long long a,long long b,long long c) { if(a<b) { long long t=a; a=b; b=t; } if(c<0) return 0; if(c<=b) return f2(c+1); if(c<=a) return f2(b+1)+f3(b+1)*(c-b)+f3(c-b)*(b+1); long long tem=f4(a,b,a); if(c<=a+b) return tem+f(a,b,a)*(c-a)+b*f3(c-a)-f2(c-a-1); return f4(a,b,a+b)+f(a,b,a+b)*(c-(a+b)); } int main() { //cout<<f(2,2,2)<<endl; //cout<<f4(3,3,6); int T; cin>>T; while(T--) { long long l1,l2,l3,l4,r1,r2,r3,r4; cin>>l1>>r1>>l2>>r2>>l3>>r3>>l4>>r4; if(r1<l1||r2<l2||r3<l3||r4<l4) { cout<<0<<endl; continue; } long long ans=0; for(long long d=l4;d<=r4;d++) { ans+=f4(r1-l1,r2-l2,r3-d-l1-l2)-f4(r1-l1,r2-l2,l3-d-l1-l2-1); ans+=f4(r2-l2,r3-l3,r1-d-l2-l3)-f4(r2-l2,r3-l3,l1-d-l2-l3-1); ans+=f4(r1-l1,r3-l3,r2-d-l1-l3)-f4(r1-l1,r3-l3,l2-d-l1-l3-1); ans+=f4(r1-l1,r2-l2,d-l1-l2-l3)-f4(r1-l1,r2-l2,d-l1-l2-r3-1); } ans=(r1-l1+1)*(r2-l2+1)*(r3-l3+1)*(r4-l4+1)-ans; cout<<ans<<endl; } /* 666 1 2 1 2 1 2 1 888 1 888 1 2 1 2 1 2 */ return 0; }
H.1:06:48(-1) solved by hl
没有负环的条件是没有一条来的路和加的边加起来为负，所以每次加的边为反向最短路
#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 = 310; const int maxm = 710; const LL INF = 1e18; const int mod = 1e9 + 7; int N,M,K; struct Edge{ int to,next; LL dis; }edge[maxm * 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v,LL w){ edge[tot].to = v; edge[tot].next = head[u]; edge[tot].dis = w; head[u] = tot++; } LL dis[1005],vis[1005]; struct node{ int pos; LL val; node(int pos,LL val):pos(pos),val(val){} friend bool operator < (node a,node b){ return a.val > b.val; } }; LL Dijkstra(int s,int t){ for(int i = 0; i <= N ; i ++) dis[i] = INF; dis[s] = 0; priority_queue<node>Q; Q.push(node(s,0)); while(!Q.empty()){ node u = Q.top(); Q.pop(); if(u.val > dis[u.pos]) continue; for(int i = head[u.pos]; ~i; i = edge[i].next){ int v = edge[i].to; if(dis[v] > edge[i].dis + u.val){ dis[v] = edge[i].dis + u.val; Q.push(node(v,dis[v])); } } } return dis[t]; } int main(){ int T = read(); while(T--){ Sca2(N,M); init(); for(int i = 1; i <= M ; i ++){ int u = read(),v = read(); LL w = read(); add(u,v,w); } for(int i = 1; i <= 6; i ++){ int u = read(),v = read(); //S = u,T = v; LL ans = -Dijkstra(v,u); add(u,v,ans); Prl(ans); } } return 0; }
I.2:40:04(-5) solved by gbs
已经多次证实两个人敲同一道题的后果就是罚时++
隐约听到了队友说的类似于玄学，暴力之类的做法
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef long long LL; int an[1123456]; int ans[123456]; int main() { int N; int y; while(cin >>N>>y) { for (int i=0; i<N; i++) { scanf("%d",&an[i]); } if (N ==1) { for (int j=1 ;j<=y; j++) { ans[j] = an[0] + j; } } else { sort(an,an+N); int nowi = N-1; int now_ans ; for (int j=y; j>=1; j--) { if (nowi == 0) { ans[j] = an[N-1]+j; continue; } now_ans = max(an[N-1]+j,an[nowi-1]+y); while(an[nowi-1] + j<=an[nowi]) { nowi--; if (nowi == 0) { now_ans = an[N-1]+j; break; } now_ans = max(an[N-1]+j,an[nowi-1]+y); } ans[j] = now_ans; } } for (int j=1; j<=y; j++) { if (j!=1) printf(" "); printf("%d",ans[j]); } printf("\n"); } #ifdef VSCode system("pause"); #endif return 0; }
