标签：style blog http color io os ar for sp
题目 ： http://acm.bnu.edu.cn/v3/problem_show.php?pid=40489
最近做到的一道蛮有意思的题目(codechef现在的题目确实很赞了)
题意 ：中文题面 (cc的一大好处就是有中文翻译，嘿嘿)
区间Max = max{a_i|gcd(a_i, g) > 1 && x <= i <= y}
做法 ： 一开始我是用分块做的，复杂的O(m * sqrt(n) * C) C 是所含不同素数的个数， C大概最大只有6或7吧
然后裸裸的T了，换了一个姿势：用线段树或者RMQ把查询从sqrt(n) -> log(n)， 这里最初的想法就是直接对于每个素数开一个线段树，但是这样预处理空间和时间都过不了
，但是考虑的其实n个数都分解一下只有C*n个数，那么的话我们可以用动态开线段树的方法，当然这个我不会- - ！，还有一种比较好的实现方式是对于每个素数都建立一段段的RMQ，额，， 具体就是比如 a[] : 2, 3, 4, 6 那么我建立一个RMQ为 ：2 4 6 3 6，前三个是2的倍数，后两个是3的倍数，，并通过一些其他辅助操作可以使得复杂度降为log(n)。
分块做法(TLE) :
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <vector> 5 6 using namespace std; 7 inline int gcd(int a, int b) {return !b ? a : gcd(b, a % b);} 8 const int MAXN = 100005; 9 int check[MAXN] = {0}; 10 const int M = 320; 11 vector<int> prime[MAXN]; 12 int ptot = 0; 13 int n, m; 14 int blocks[1<<9][10005]; 15 int a[MAXN]; 16 void init(int N) { 17 for (int i = 2; i <= N; i++) { 18 if (!check[i]) { 19 for (int j = 1; j * i <= N; j++) { 20 prime[j * i].push_back(i); 21 check[i * j] = 1; 22 } 23 check[i] = ++ ptot; 24 } 25 } 26 } 27 vector<int> G[MAXN]; 28 int query(int x, int y, int e) { 29 return upper_bound(G[e].begin(), G[e].end(), y) - lower_bound(G[e].begin(), G[e].end(), x); 30 } 31 32 int main() { 33 init(100000); 34 scanf("%d%d", &n, &m); 35 for (int i = 1; i <= n; i++) { 36 scanf("%d", &a[i]); 37 int e = i / M; 38 for (int j = 0; j < prime[a[i]].size(); j++) { 39 int v = prime[a[i]][j]; 40 blocks[e][check[v]] = max(blocks[e][check[v]], a[i]); 41 } 42 G[a[i]].push_back(i); 43 } 44 while (m --) { 45 int g, x, y; 46 scanf("%d%d%d", &g, &x, &y); 47 int l = x / M, r = y / M, Max = -1; 48 if (l == r) { 49 for (int i = x; i <= y; i++) { 50 if (gcd(g, a[i]) > 1) Max = max(Max, a[i]); 51 } 52 }else { 53 for (int i = x; i < (l+1) * M; i++) { 54 if (gcd(g, a[i]) > 1) Max = max(Max, a[i]); 55 } 56 for (int i = r * M; i <= y; i++) { 57 if (gcd(g, a[i]) > 1) Max = max(Max, a[i]); 58 } 59 for (int i = 0; i < prime[g].size(); i++) { 60 int v = prime[g][i]; 61 for (int j = l + 1; j <= r - 1; j++) { 62 if (blocks[j][v] > 0)Max = max(Max, blocks[j][v]); 63 } 64 } 65 } 66 if (Max == -1) printf("-1 -1\n"); 67 else { 68 printf("%d %d\n", Max, query(x, y, Max)); 69 } 70 } 71 return 0; 72 }
RMQ做法:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <vector> 5 6 using namespace std; 7 const int MAXN = 100005; 8 struct RMQ { 9 int a[MAXN * 7][20]; 10 int log2[MAXN * 7]; 11 void init(int A[], int N) { 12 for (int i = 0; i < MAXN * 7; i++) log2[i] = (i == 0 ? -1 : log2[i >> 1] + 1); 13 for (int i = 1; i <= N; i++) a[i][0] = A[i]; 14 for (int j = 1; (1 << j) <= N; j++) { 15 for (int i = 1; i + (1<<j) - 1 <= N; i++) { 16 a[i][j] = max(a[i][j-1], a[i + (1<<(j-1))][j - 1]); 17 } 18 } 19 } 20 int rmq(int L, int R) { 21 if (R < L) return 0; 22 int k = log2[R - L + 1]; 23 return max(a[L][k], a[R-(1<<k)+1][k]); 24 } 25 }AC; 26 vector<int> prime[MAXN]; 27 int check[MAXN] = {0}, ptot = 0; 28 void init(int N) { 29 for (int i = 2; i <= N; i++) { 30 if (!check[i]) { 31 for (int j = 1; j * i <= N; j++) { 32 prime[j * i].push_back(i); 33 check[i * j] = 1; 34 } 35 check[i] = ++ ptot; 36 } 37 } 38 } 39 vector<int> G[MAXN]; 40 int query(int x, int y, int e) { 41 return upper_bound(G[e].begin(), G[e].end(), y) - lower_bound(G[e].begin(), G[e].end(), x); 42 } 43 int n, m; 44 int b[MAXN * 7]; 45 int a[MAXN]; 46 int pre[MAXN]; 47 vector<int> gao[MAXN]; 48 49 int main() { 50 init(100000); 51 scanf("%d%d", &n, &m); 52 for (int i = 1; i <= n; i++) { 53 scanf("%d", &a[i]); 54 for (int j = 0; j < prime[a[i]].size(); j++) { 55 int v = check[prime[a[i]][j]]; 56 gao[v].push_back(i); 57 } 58 G[a[i]].push_back(i); 59 } 60 int tot = 0; 61 for (int i = 1; i <= ptot; i++) { 62 if (gao[i].size() == 0) continue; 63 pre[i] = tot; 64 for (int j = 0; j < gao[i].size(); j++) { 65 b[++ tot] = a[gao[i][j]]; 66 } 67 } 68 AC.init(b, tot); 69 while (m--) { 70 int g, x, y, Max = -1; 71 scanf("%d%d%d", &g, &x, &y); 72 for (int i = 0; i < prime[g].size(); i++) { 73 int v = check[prime[g][i]]; 74 int r = upper_bound(gao[v].begin(), gao[v].end(), y) - gao[v].begin() + pre[v] + 1; 75 int l = lower_bound(gao[v].begin(), gao[v].end(), x) - gao[v].begin() + pre[v] + 1; 76 r --; 77 if (r >= l)Max = max(Max, AC.rmq(l, r)); 78 } 79 if (Max == -1) { 80 printf("-1 -1\n"); 81 }else { 82 printf("%d %d\n", Max, query(x, y, Max)); 83 } 84 } 85 return 0; 86 }
Maximum number, GCD condition (codechef March Challenge 2014)
标签：style blog http color io os ar for sp
原文地址：http://www.cnblogs.com/danceonly/p/4029818.html
