数学 之 Codeforces 359D - Pair of Numbers

//  [7/13/2014 Sjm]/*直接暴力，超时。。不过有一点大家都知道：如果 （b%a == 0），(c%b == 0)， 那么 （c%a == 0） 一定是成立的。 故而在以一个数字a为中心，向两边寻找能够被a整除的数后，已被寻找到的的数是不用再计算的。。。(因为即使计算这些已被寻找到的数，边界r、l也一定会 小于或等于 以数字a为中心所得到的边界r、l。) ，，，比赛时怎么就没想到呢。。。。*/
#include <iostream>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;const int MAX = 300005;int arr[MAX], arr_L[MAX];int n;void Solve(){int cnt = 0, dis = -1;int l, r;for (int i = 0; i < n;) {l = r = i;while (l && !(arr[l - 1] % arr[i])) { --l; }while (r < (n - 1) && !(arr[r + 1] % arr[i])) { ++r; }i = r + 1;int t_dis = r - l;if (t_dis > dis) {cnt = 0;dis = t_dis;}if (t_dis == dis) {arr_L[cnt++] = l + 1;}}printf("%d %d\n", cnt, dis);for (int i = 0; i < cnt; i++) {printf("%d ", arr_L[i]);}}int main(){scanf("%d", &n);for (int i = 0; i < n; ++i) {scanf("%d", &arr[i]);}Solve();return  0;}

//  [7/11/2014 Sjm]/*这是另外一种做法：虽然有些麻烦，但收获很多（Sparce Table算法，以及又一次二分（话说对二分有阴影）） 知识点：math + Sparce Table算法（DP） + 二分 math：数组中[l, r]区间中任意一个数都能被aj整除，则aj必然满足aj是[l,r]中最小的。。即： 对于区间[l, r]， min(l, r) = gcd(l, r) Sparce Table算法（DP）：为了在O(1)的时间，获取某区间的min以及gcd，采用Sparce Table算法。Sparce Table算法：（本质就是DP）预处理：状态：dp[i][j] := 区间[i, i+2^j-1]的函数F值决策：dp[i][j] = F(dp[i][j-1],  dp[i+2^(j-1)][j-1])（此题中 函数F 即：min(), gcd()）查询：设查询到区间为 [m, n]，区间总共有 (n-m+1) 个数据方程 2^k <= (n-m+1) 求解出 k，可得：dp[m][n] = F(dp[m][k], dp[n-2^k+1][k])（表达式数中有重叠，但保证了结果的正确性） 二分：采用二分的方法，寻找出满足条件的 (r-l) 的最大值*/
#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>using namespace std;const int MAX_num = 300005;const int MAX_pow = 20;const int INF = 0x3f3f3f3f;int T, n;int arr_min[MAX_num][MAX_pow];int arr_gcd[MAX_num][MAX_pow];int Gcd(int a, int b) {if (0 == b) { return a; }else return Gcd(b, a%b);}void ST() {int MAX_j = (int)(log2((double)(n))) + 1;for (int j = 1; j < MAX_j; ++j) {for (int i = 0; i < n; ++i) {int tep = 1 << (j - 1);if (i + tep < n) {arr_min[i][j] = min(arr_min[i][j - 1], arr_min[i + tep][j - 1]);arr_gcd[i][j] = Gcd(arr_gcd[i][j - 1], arr_gcd[i + tep][j - 1]);}}}}bool Judge(int len) {int tep = (int)(log2((double)(len + 1)));for (int i = 0; i < n - len; i++) {int j = i + len;int t_min = min(arr_min[i][tep], arr_min[j - (1 << tep) + 1][tep]);int t_gcd = Gcd(arr_gcd[i][tep], arr_gcd[j - (1 << tep) + 1][tep]);if (t_gcd == t_min) {return true;}}return false;}int Binary_search() {int l = 0, r = n - 1;int mid;while (l < r) {mid = l + ((r - l + 1) >> 1);//cout << "l = " << l << endl;if (Judge(mid)) { l = mid; }else { r = mid - 1; }}//cout << "l = " << l << endl;return l;}void myOutput(int len) {vector<int> vec;int tep = (int)(log2((double)(len + 1)));if (len > 0) {for (int i = 0; i < n - len; i++) {int j = i + len;int t_min = min(arr_min[i][tep], arr_min[j - (1 << tep) + 1][tep]);int t_gcd = Gcd(arr_gcd[i][tep], arr_gcd[j - (1 << tep) + 1][tep]);if (t_gcd == t_min) {vec.push_back(i);}}}else {for (int i = 0; i < n; ++i) {vec.push_back(i);}}printf("%d %d\n", vec.size(), (len > 0) ? len : 0);for (int i = 0; i < vec.size(); ++i) {printf("%d ", vec[i] + 1);}printf("\n");}int main(){//freopen("input.txt", "r", stdin);memset(arr_min, INF, sizeof(arr_min));memset(arr_gcd, INF, sizeof(arr_gcd));scanf("%d", &n);for (int i = 0; i < n; ++i) {scanf("%d", &arr_min[i][0]);arr_gcd[i][0] = arr_min[i][0];}ST();myOutput(Binary_search());return 0;}