CodeForces 359D Pair of Numbers (暴力)

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;
}