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 a_i mod a_j=0,now OO want to know

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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 a_i(0<a_i<=10000)

 

Output

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

 

Sample Input

51 2 3 4 5

 

Sample Output

23

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

解题思路：
预处理出所有数的因数并且保存每个数出现的位置，对于每个A[i]， 找最小的区间使得该区间内不包含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;}