div1真难，现在就是后悔， 非常后悔
A.显然如果我们知道起点和终点是哪两个点，我们可以算出距离通过b / gcd(a,b)的方式求出需要走几步的。
并且数据范围似乎也允许我们这么做，所以直接枚举取最大小值就可以了
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #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 ", x) #define Prl(x) printf("%lld ",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 eps = 1e-9; const int maxn = 110; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; LL N,K,s,t; LL gcd(LL a,LL b){ return b == 0?a:gcd(b,a % b); } LL lcm(LL a,LL b){ return b / gcd(a,b); } LL check(LL s,LL t){ if(t < 1 || t > N * K) return -1; if(s < 1 || s > N * K) return -1; LL d = t - s; while(d <= 0) d += N * K; while(d > N * K) d -= N * K; return lcm(d,N * K); } LL solve(LL S){ if(S > N * K || S < 1) return 1e18; LL ans = 1e18; for(int i = 0; i < N ; i ++){ LL x = check(S,i * K + 1 + t); if(~x) ans = min(ans,x); x = check(S,(i + 1) * K + 1 - t); if(~x) ans = min(ans,x); } return ans; } LL solve2(LL S){ if(S > N * K || S < 1) return 0; LL ans = 0; for(int i = 0; i < N ; i ++){ LL x = check(S,i * K + 1 + t); if(~x) ans = max(ans,x); x = check(S,(i + 1) * K + 1 - t); if(~x) ans = max(ans,x); } return ans; } int main(){ scanf("%lld%lld%lld%lld",&N,&K,&s,&t); LL ans = solve(s + 1); ans = min(ans,solve(K + 1 - s)); printf("%lld ",ans); ans = solve2(s + 1); ans = max(solve2(K + 1 - s),ans); printf("%lld",ans); return 0; }
B.显然第一个全排列类似于题目重定义了一个全排列的顺序，那我们依照这个像字典序一样的东西，把需要查询的排列换回正常的序列。
题目就变成了区间查询一个序列内是否存在1 - N的全排列的环的问题。
事实上不难发现，每一个点如果要跳的话贪心的跳最近的下一个点是最好的，跳完整个全排列就是跳N - 1步的路程。
对于查询的预处理，想到的是预处理出一个数字r[i]表示i这个点为起点最近的终点使得之间包含一个全排列环
那么首先可以想到i点跳N - 1步的位置就是r[i],对于每个点，倍增一下就可以了，时间复杂度O(nlogn)
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #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 ", x) #define Prl(x) printf("%lld ",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 eps = 1e-9; const int maxn = 2e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,Q; int Hash[maxn]; int b[maxn],near[maxn]; int fa[maxn],dp[maxn]; int nxt[maxn]; vector<int>Stack[maxn]; int find(int t){ if(t == fa[t]) return t; return fa[t] = find(fa[t]); } const int SP = 20; int pa[maxn][SP]; int r[maxn]; int main(){ Sca3(N,M,Q); for(int i = 1; i <= N; i ++) Hash[read()] = i; for(int i = 1; i <= M; i ++) b[i] = Hash[read()]; for(int i = M; i >= 1; i --){ int la = b[i] + 1; if(la == N + 1) la = 1; if(!near[la]) nxt[i] = M + 1; else nxt[i] = near[la]; near[b[i]] = i; } for(int i = 1; i <= M ; i ++) pa[i][0] = nxt[i]; pa[M + 1][0] = M + 1; for(int i = 1; i < SP; i ++){ for(int j = 1; j <= M + 1; j ++){ pa[j][i] = pa[pa[j][i - 1]][i - 1]; } } //cout << pa[1][0] << endl; N--; for(int i = 1; i <= M ; i ++){ r[i] = i; for(int j = SP - 1; j >= 0; j --){ if(N & (1 << j)) r[i] = pa[r[i]][j]; } } for(int i = M - 1; i >= 1; i --) r[i] = min(r[i],r[i + 1]); while(Q--){ int L,R; Sca2(L,R); if(R < r[L]) printf("0"); else printf("1"); } return 0; }
事实上还是有O(n)的做法的，如果将每一个点往他的前驱连边，就会发现形成一颗树，节点t的答案就是往上数N个祖先。
这就可以线性做出来了。
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #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 ", x) #define Prl(x) printf("%lld ",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 eps = 1e-9; const int maxn = 2e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,Q; int Hash[maxn]; int b[maxn],near[maxn]; int fa[maxn],dp[maxn]; int nxt[maxn]; vector<int>Stack[maxn]; int find(int t){ if(t == fa[t]) return t; return fa[t] = find(fa[t]); } struct Edge{ int to,next; }edge[maxn * 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= M + 1; i ++) head[i] = -1; tot = 0; } void add(int u,int v){ // cout << u << ' ' << v << endl; edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } int r[maxn]; int pos[maxn],dep[maxn]; void dfs(int t){ pos[dep[t]] = t; if(dep[t] >= N) r[t] = pos[dep[t] - N + 1]; else r[t] = M + 1; for(int i = head[t]; ~i; i = edge[i].next){ int v = edge[i].to; dep[v] = dep[t] + 1; dfs(v); } } int main(){ Sca3(N,M,Q); init(); for(int i = 1; i <= N; i ++) Hash[read()] = i; for(int i = 1; i <= M; i ++) b[i] = Hash[read()]; for(int i = M; i >= 1; i --){ int la = b[i] + 1; if(la == N + 1) la = 1; if(!near[la]) nxt[i] = M + 1; else nxt[i] = near[la]; near[b[i]] = i; } for(int i = 1; i <= M ; i ++) add(nxt[i],i); dep[M + 1] = 0; dfs(M + 1); for(int i = M - 1; i >= 1; i --) r[i] = min(r[i],r[i + 1]); while(Q--){ int L,R; Sca2(L,R); if(R < r[L]) printf("0"); else printf("1"); } return 0; }