题意:给定一个正整数数组,求最长的区间,使得该区间内存在一个元素,它能整除该区间的每个元素。
析:暴力每一个可能的区间,从数组的第一个元素开始考虑,向两边延伸,设延伸到的最左边的点为l, 最右边的点为r。那么我们下一点考虑r+1即可,
因为[l, r]之间不会有更优解。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<double, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-5; const int maxn = 3e5 + 10; const int mod = 1e6; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[maxn]; int main(){ scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%d", a+i); int r = 0; int ans = 0; vector<int> v; for(int i = 1; i <= n; i = r){ if(a[i] == 1){ printf("1 %d 1 ", n-1); return 0; } int l = i; r = i; while(a[l] % a[i] == 0 && l > 0) --l; while(r <= n && a[r] % a[i] == 0) ++r; if(ans < r-l-2){ ans = r-l-2; v.clear(); } if(ans == r-l-2) v.push_back(l+1); } printf("%d %d ", v.size(), ans); for(int i = 0; i < v.size(); ++i){ if(i) putchar(' '); printf("%d", v[i]); } printf(" "); return 0; }