HDU 5288 OO's sequence (2015多校第一场 二分查找）

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory
Limit: 131072/131072 K (Java/Others)

Total Submission(s): 955 Accepted Submission(s): 358



Problem Description

OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know










































（









）


Input

There are multiple test cases. Please process till EOF.

In each test case:

First line: an integer n(n<=10^5) indicating the size of array

Second line:contain n numbers ai(0<ai<=10000)

Output

For each tests: ouput a line contain a number ans.

Sample Input

5
1 2 3 4 5

Sample Output

23

Author

FZUACM

Source

2015 Multi-University Training Contest 1
解题思路：

预处理出全部数的因数而且保存每一个数出现的位置，对于每一个A[i]， 找最小的区间使得该区间内不包括A[i]的因数，通过向左向右进行两次二分查找来实现。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
const int MAXN = 100000 + 10;
const int MOD = 1e9 + 7;
vector<int>G[MAXN];
vector<int>p[MAXN];
vector<int>rp[MAXN];
vector<int>M[MAXN];
int N;
int A[MAXN];
int main()
{
for(int i=1; i<=10000; i++)
{
for(int j=i; j<=10000; j+=i)
G[j].push_back(i);
}
while(scanf("%d", &N)!=EOF)
{
for(int i=1;i<=N;i++) p[i].clear();
for(int i=1;i<=N;i++) rp[i].clear();
for(int i=1; i<=N; i++)
scanf("%d", &A[i]);
for(int i=1; i<=N; i++)
{
p[A[i]].push_back(i);
}
for(int i=N;i>=1;i--)
{
rp[A[i]].push_back(-i);
}
long long ans = 0;
for(int i=1; i<=N; i++)
{
int L = -1, R = N + 1;
for(int j=0; j<G[A[i]].size(); j++)
{
int x = G[A[i]][j];
int r = upper_bound(p[x].begin(), p[x].end(), i) - p[x].begin();
int l = upper_bound(rp[x].begin(), rp[x].end(), -i) - rp[x].begin();
if(r >= p[x].size()) r = N  + 1;
else r = p[x][r];
//cout << x << ": " << l << ' '<< r << endl;
if(l >= rp[x].size()) l = 0;
else l = -(rp[x][l]);
//cout << x << ": " << l << ' '<< r << endl;
L = max(l, L);
R = min(R, r);
}
if(R == 0) R = N + 1;
if(L < 0) L = 0;
//cout << L << ' ' << R << endl;
ans = (ans + (long long)(i - L) * (long long)(R - i)) % MOD;
}
printf("%I64d\n", ans );
}
return 0;
}