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  • 2015多校联赛 ——HDU5288(数学)

    OO’s Sequence

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 449    Accepted Submission(s): 158

    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
    i=1nj=inf(i,j) mod 10^9+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 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
     
    Source

    求在[a , b] 中有多少个i 满足找不到另一个数j 使,a[i] % a[j] = 0
    用L[],R[]分别记录左右离起最近的因子,ans+=(R[i]-i+1)*(i-L[i]+1)/2 %MOD
    //自己并没有想到怎么做 (╯▽╰)

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    using namespace std ;
    const int maxn = 100010 ;
    const int mod = 1e9+7 ;
    int a[maxn];
    long long  last[maxn] ;
    long long pre[maxn] ;
    long long  l[maxn] ;
    long long r[maxn] ;
    int  main()
    {
        //freopen("1001.txt" ,"r" ,stdin) ;
        int n;
        while(~scanf("%d",&n))
        {
            memset(pre,0,sizeof(pre));
            memset(last,0,sizeof(last));
            for(int i = 1;i <= n;i++)
            {
                l[i] = 1;
                r[i] = n;
                scanf("%d",&a[i]);
                for(int j = a[i];j < 10001;j+=a[i])
                {
                    if(pre[j] && r[pre[j]] == n)
                        r[pre[j]] = i-1;
                    pre[a[i]] = i;
                }
            }
    
    
            for(int i = n;i >=1;i--)
            {
                for(int j = a[i];j < 10001;j+=a[i])
                {
                    if(last[j] && l[last[j]]==1)
                        l[last[j]] = i+1;
                    last[a[i]] = i;
                }
            }
            long long ans = 0;
            for(long long int i = 1;i <= n;i++)
                ans= (ans+(((i-l[i]+1)%mod)*((r[i] - i +1)%mod))%mod)%mod;
            printf("%I64d
    ",ans);
        }
        return 0 ;
    }



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  • 原文地址:https://www.cnblogs.com/Przz/p/5409832.html
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