2018ICPC 南京 训练赛
赛后总结：
1.补字符串
2.前期交题看三遍
3.认真听取zcz兄弟的意见 最后时间该冲的时候就要冲
A.分情况讨论一下 K > 2先手必胜
N可能是0
solved by gbs
00:17:41(-2) 没有考虑0 WA2发
#include <iostream> #include<stack> #include<math.h> #include<stdlib.h> #include<string.h> #include<string> #include<ctime> #include<complex> #include<stdio.h> #include<algorithm> #include<map> #include<deque> using namespace std; typedef long long LL; int main() { int n,k; while(cin >>n >>k) { if (n == 0) { printf("Austin\n"); } else if (k ==1) { if (n&1) printf("Adrien\n"); else printf("Austin\n"); } else { printf("Adrien\n"); } } return 0; }
D.经典的最小球覆盖问题，方法一般分为三分套三分套三分或者随机算法
网络上有很多类似的解法，就不赘述了
solved by hl
0:41:43
#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 = 110; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; struct Point{ double x,y,z; Point(){} Point(double x,double y,double z):x(x),y(y),z(z){} }point[maxn],ans; double A; double ansx,ansy; const double delta = 0.999; double check(double x,double y,double z){ double t = 0; for(int i = 1; i <= N ; i ++){ double d = (x - point[i].x) * (x - point[i].x) + (y - point[i].y) * (y - point[i].y) + (z - point[i].z) * (z - point[i].z); d = sqrt(d); t = max(d,t); } return t; } void sa(){ double t = 3000; Point a = ans; while(t > 1e-18){ Point anstmp = a; anstmp.x += (rand() * 2 - RAND_MAX) * t; anstmp.y += (rand() * 2 - RAND_MAX) * t; anstmp.z += (rand() * 2 - RAND_MAX) * t; double newans = check(anstmp.x,anstmp.y,anstmp.z); double DE = newans - A; if(DE < 0){ ans = a = anstmp; A = newans; }else if(exp(-DE / t) * RAND_MAX > rand()){ a = anstmp; } t = t * delta; } } void SA(){ sa(); sa(); sa(); sa(); sa(); sa(); } int main(){ Sca(N); srand(time(NULL)); for(int i = 1; i <= N ; i ++){ scanf("%lf%lf%lf",&point[i].x,&point[i].y,&point[i].z); ans.x += point[i].x; ans.y += point[i].y; ans.z += point[i].z; } ans.x /= N; ans.y /= N; ans.z /= N; A = check(ans.x,ans.y,ans.z); //cout << A << endl; //cout << ans.x << " " << ans.y << " " << ans.z <<endl; SA(); printf("%.15lf",A); return 0; }
G.数学功底强的直接把式子推出来
不想推式子的证明他的式子是一个4次以下的多项式 然后拉格朗日插值
solved by gbs
1:49:51(-3)
参数没调好TLE + RE
#include <iostream> #include<stack> #include<math.h> #include<stdlib.h> #include<string.h> #include<string> #include<ctime> #include<complex> #include<stdio.h> #include<algorithm> #include<map> #include<deque> using namespace std; typedef long long LL; const int mod =1e9+7; LL ans ; LL qpow(LL a,int k){ LL ans =1; while(k!=0) { if (k&1)ans =(ans*a)%mod; a= (a*a)%mod; k/=2; } return ans; } int xn[6]; int yna[6]; int hn[6]; int sizea= 5; int main() { //LL inv6=qpow(6,mod-2); int inv[45]; yna[1] =1; yna[2] = 5; yna[3] = 15; yna[4] = 35; yna[5] = 70; for (int i=1; i<=35; i++) inv[i] =qpow(i-10,mod-2); int t; int n; cin>>t; for (int i=1; i<=5; i++) { LL llans = 1; for (int j =1; j<=5; j++) { if (j == i)continue; llans = ((llans *inv[i-j+10])%mod+mod)%mod; } hn[i] = llans; } while(t--) { scanf("%d",&n); LL ans = 0; for (int i=1; i<=5; i++) { LL llans = 1; for (int j =1; j<=5; j++) { if (j == i)continue; llans = ((llans *(n-j))%mod+mod)%mod; } llans =(llans*hn[i])%mod; //cout<<llans<<endl; ans = (ans+llans*yna[i])%mod; } printf("%lld\n",ans); /*ans = n; ans =(ans*(n+1))%mod; ans =(ans*(2LL*n+1))%mod; ans =(ans*i)%mod;*/ } return 0; }
I.似乎是个最大流模型
不过似乎也有非网络流做法。
将每个hero对每个monster连接一条容量为1的边，然后源点对每个hero连接一条容量为1的边，monster对汇点连1的边。
然后源点连出一个点作为魔药，容量为K，魔药点对每个hero连接一条容量为1的边
solved by hl
1:04:51
#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 = 1210; const int maxm = 1e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; struct Dinic{ struct Edge{ int from,to,next,cap,flow; Edge(){} Edge(int from,int to,int next,int cap,int flow):from(from),to(to),next(next),cap(cap),flow(flow){} }edge[maxm * 2]; int n,s,t,head[maxn],tot; int dep[maxn],cur[maxn]; void init(int n,int s,int t){ this->n = n; this->s = s; this->t = t; tot = 0; for(int i = 0 ; i <= n ; i ++) head[i] = -1; } inline void add(int s,int t,int w){ // cout << s << " " << t << " " << w <<endl; edge[tot] = Edge(s,t,head[s],w,0); head[s] = tot++; edge[tot] = Edge(t,s,head[t],0,0); head[t] = tot++; } inline bool BFS(){ for(int i = 0 ; i <= n ; i ++) dep[i] = -1; dep[s] = 1; queue<int>Q; Q.push(s); while(!Q.empty()){ int u = Q.front(); Q.pop(); for(int i = head[u]; ~i ; i = edge[i].next){ int v = edge[i].to; //cout <<edge[i].flow << " " << edge[i] //cout << v << " " << dep[v] << endl; if(~dep[v] || edge[i].flow >= edge[i].cap) continue; dep[v] = dep[u] + 1; Q.push(v); } } return ~dep[t]; } inline int DFS(const int& u,int a){ if(u == t || !a) return a; int flow = 0; for(int &i = cur[u]; ~i; i = edge[i].next){ int v = edge[i].to; if(dep[v] != dep[u] +1) continue; int f = DFS(v,min(a,edge[i].cap - edge[i].flow)); if(!f) continue; edge[i ^ 1].flow -= f; edge[i].flow += f; a -= f; flow += f; } return flow; } inline int maxflow(){ return maxflow(s,t); } inline int maxflow(int s,int t){ int flow = 0; while(BFS()){ //cout << "bug" <<endl; for(int i = 0 ; i <= n ; i ++) cur[i] = head[i]; flow += DFS(s,INF); } return flow; } }g; int main(){ Sca3(N,M,K); int S= N + M + 3,T = N + M + 4; g.init(T,S,T); g.add(S,N + M + 1,INF); g.add(S,N + M + 2,K); for(int i = 1; i <= N ; i ++){ g.add(N + M + 1,M + i,1); g.add(N + M + 2,M + i,1); } for(int i = 1; i <= M ; i ++) g.add(i,T,1); for(int i = 1; i <= N ; i ++){ int k = read(); while(k--){ int x = read(); g.add(M + i,x,1); } } Pri(g.maxflow()); return 0; }
J.
当仅考虑素数的时候，当一个点i含素数因子时，对答案产生的贡献是(i + 1) * (N - i + 1),即包括这个点的所有范围都会出现这个素数，每个范围产生1的贡献，总贡献就是范围的个数
所以说我们用素数筛预处理出每个数含有的所有素数因子，对于每一个素数因子，出现的时候都会产生以上的贡献。
剩下的就是考虑出现相同的素数重复计算的部分了
形如 2 1 1 2 1 1 2 1 1 1的样例
素数2的下标为1 4 7
素数2产生的贡献应当是(1 - 0) * (10 - 1 + 1) + (4 - 1) * (10 - 4 + 1) + (7 - 4) * (10 - 7 + 1)
所以还需要一个数组记录每个素数当前出现的最右端的位置
solved by hl
0:25:30 (-1)
运算过程中爆LL WA一发
#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,K; int a[maxn]; vector<int>prime[maxn]; int isprime[maxn]; void init(){ for(int i = 2; i < maxn; i ++) isprime[i] = 1; for(int i = 2; i < maxn; i ++){ if(!isprime[i]) continue; prime[i].pb(i); for(int j = i + i; j < maxn; j += i){ isprime[j] = 0; prime[j].pb(i); } } } LL pos[maxn]; int main(){ Sca(N); init(); LL ans = 0; for(LL i = 1; i <= N ; i ++){ Sca(a[i]); for(int j = 0 ; j < prime[a[i]].size(); j ++){ int v = prime[a[i]][j]; ans += 1ll * (i - pos[v]) * (N - i + 1); pos[v] = i; } } Prl(ans); return 0; }
K.
用脑子想策略 (×)
随机盲一发 (√)
在这些奇怪的类似构造题上 就要想一些奇怪的方法。
unsolved by gbs
M.
应该是manacher计算回文串的贡献 + 计算LCP
写了AC自动机最后发现爆时间复杂度
unsolved by hl
