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  • hdu5288(OO’s Sequence)

    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 imod a j=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 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

    题意:给定一个正整数序列a,f(l,r)表示在a[l...r]中的a[i],对于a[l...r]中的每一个数a[j]都有a[i]%a[j]!=0的i数目(i!=j)。最后求:



    题解:稍加思考很简单的一道题,设l[i]、r[i]分别表示a[i]左右最近的约数的下标,不存在则分别为-1和n。注意到a[i]<10000,所以可以用一个标记数组flag来标记a[i]某个约数是否出现过。最终答案为sigma[(i-l[i])*(r[i]-i)]mod(1e10+7)


    #include <iostream>
    #include <sstream>
    #include <fstream>
    #include <string>
    #include <map>
    #include <vector>
    #include <list>
    #include <set>
    #include <stack>
    #include <queue>
    #include <deque>
    #include <algorithm>
    #include <functional>
    #include <iomanip>
    #include <limits>
    #include <new>
    #include <utility>
    #include <iterator>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cctype>
    #include <cmath>
    #include <ctime>
    using namespace std;
    
    int a[100010], l[100010], r[100010], flag[10010];
    
    int main()
    {
        int n;
        while (~scanf("%d", &n))
        {
            for (int i = 0; i < n; ++i)
                scanf("%d", &a[i]);
            fill(l, l+n, -1);
            fill(r, r+n, n);
            //计算a[i]左端最近的约数
            memset(flag, -1, sizeof(flag));
            for (int i = 0; i < n; ++i)
            {
                for (int j = 1; j*j <= a[i]; ++j)
                    if (a[i] % j == 0)
                    {
                        if (flag[j] != -1)
                            l[i] = max(l[i], flag[j]);
                        if (flag[a[i]/j] != -1)
                            l[i] = max(l[i], flag[a[i]/j]);
                    }
                flag[a[i]] = i;
            }
            //计算a[i]右端最近的约数
            memset(flag, -1, sizeof(flag));
            for (int i = n-1; i >= 0; --i)
            {
                for (int j = 1; j*j <= a[i]; ++j)
                    if (a[i] % j == 0)
                    {
                        if (flag[j] != -1)
                            r[i] = min(r[i], flag[j]);
                        if (flag[a[i]/j] != -1)
                            r[i] = min(r[i], flag[a[i]/j]);
                    }
                flag[a[i]] = i;
            }
    
            int ans = 0;
            for (int i = 0; i < n; ++i)
                ans = (ans + (i - l[i]) * (r[i] - i)) % 1000000007;
            printf("%d
    ", ans);
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/godweiyang/p/12203969.html
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