Hdu 5288 OO’s Sequence 2015多小联赛A题

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1751    Accepted Submission(s): 632

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

∑i=1n∑j=inf(i,j) mod （109+7）.

 

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

5
1 2 3 4 5

 

Sample Output

23

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

 

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题意非常easy ：求随意区间内满足条件的ai有多少个

假设依照题意 程序应该这样写：

四层循环 - - （不超时吃键盘）

尽管最后优化到 n^2/2也超时 数据太大了

#include<stdio.h>
int main()
{
    int n,i,j,k,kk,a[100050];
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1; i<=n; i++)
            scanf("%d",&a[i]);
        long long suma=0;
        for(i=1; i<=n; i++)
        {
            for(j=i; j<=n; j++)
            {

                for(k=i; k<=j; k++)//相当于i
                {
                    int flag=0;
                    for(kk=i; kk <= j ; kk++)//相当于j
                    {
                        if(a[k]%a[kk] == 0 && k!=kk)
                        {
                            flag=1;
                            break;
                        }
                    }
                    if(flag!=1)
                    {
                        suma++;
                        suma%=100000007;
                    }
                }
                printf("%d %d=%d %d=%d\n",i,j,a[i],a[j],suma);
            }
        }
        printf("%I64d\n",suma);
    }
}

脑洞大开：换个思路是不是题意求的是找那些区间能满足第ai个值存在呢？

也就是说看ai能提供几个答案 

定义两个数组 l r 表示i数的左側和右側
最接近他的值且值是a[i]因子的数字的位置
那么第i个数字能提供的答案就是(r[i]-i) * (l[i]-i)
事实上这样的方法有漏洞 假设我给你 10w个1 程序就跪了 ╮(╯▽╰)╭  没有这数据 所以放心大胆的做吧

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int M = 10e5 + 5;
const long mod = 1e9+7;
int vis[M],a[M],l[M],r[M];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
        memset(vis,0,sizeof(vis));

        for(int i = 1;i <= n; ++i)
        {
            scanf("%d",&a[i]);
            r[i] = n+1;
            for(int j = a[i];j <= 10000; j+=a[i]) //找到离他近期的因子
            {
                if(vis[j])
                {
                    r[vis[j]] = i;
                    vis[j] = 0;
                }
            }
            vis[a[i]] = i;
        }
        memset(vis,0,sizeof(vis));
        for(int i = n;i >= 1; --i)
        {
            for(int j = a[i];j <= 10000; j+=a[i])
            {
                if(vis[j])
                {
                    l[vis[j]] = i;
                    vis[j] = 0;
                }
            }
            vis[a[i]] = i;
        }

        long long ans = 0;
        for(int i = 1;i <= n; ++i)
        {
        ans = ((ans + (long long)(r[i]-i)*(long long)(i-l[i])%mod)%mod);
        }

        printf("%I64d\n",ans);
    }
}