HDU 5289 OO’s Sequence(2015多校联合)

HDUOJ 题解

题目链接：戳我

题目大意：

f(l, r)代表一个l，r区间内，符合任选i，j （l <= j <= r && j != i），有$a_{i}$ % $a_{j}$ != 0的i的个数
求 $\sum_{i = 1}^n \sum_{j = i}^n f(i, j) mod(10^9+7)$

样例解释

5 即为n，代表有n个数，n <= $10^5$
1 2 3 4 5 分别为上面的 n 个数，0 < $a_{i}$ <= 10000
f(1,1) = {1} = 1; f(1,2) = {1} = 1; f(1,3) = {1} = 1; f(1,4) = {1} = 1; f(1,5) = {1} = 1
f(2,2) = {2} =1; f(2,3) = {2,3} = 2; f(2,4) = {2,3} = 2; f(2,5) = {2,3,5} = 3
f(3,3) = {3} = 1; f(3,4) = {3,4} = 2; f(3,5） = {3,4,5} = 3；
f(4,4) = {4} = 1; f(4, 5) = {4,5} = 2;
f(5,5) = {5} = 1;
故所有加起来为23

解题思路

分析：数组长度为$10^5$,而每个数不超多 10000
假设一个数的位置是 pos 值为x，那个在x左侧和x最近的属于x的因子的位置是l，在x右侧和x最近的属于x因子的位置是r，那么在(l+1， r-1)这个区间内，任选左区间一个位置ll，在任选右区间一个位置rr，那么x必为此区间符合条件的一个数
故 $num = (x-l)*(r-x)$
只需枚举每一个位置，极其l，r即可

思路：

可以先预处理10000个数的每个数的因子，在枚举每一个位置的时候即可知道最左的因子和最右的因子了

代码：

452ms

//Author LJH
//www.cnblogs.com/tenlee
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#define clc(a, b) memset(a, b, sizeof(a))
using namespace std;
const int inf = 0x3f;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5+5;
const long long mod = 1e9+7;
int a[maxn], ha[maxn], l[maxn], r[maxn], pre[maxn];
vector<int> v[maxn];
void Init()
{
    v[1].push_back(1);
    for(int i = 2; i <= maxn; i++)
    {
        for(int j = 2; i*j <= maxn; j++)
        {
           v[j*i].push_back(i); 
        }
        v[i].push_back(1);
        v[i].push_back(i);
    }
}
int main()
{
    //freopen("1001.in", "r", stdin);
    int n, i, j, k = 0;
    long long ans;
    Init();
    while(~scanf("%d", &n))
    {
        for(i = 1; i <= n; i++)       
        {
            scanf("%d", &a[i]);
        }
        clc(pre, 0);
        for(i = 1; i <= n; i++)
        {
            l[i] = 1;
            k = a[i];
            for(j = 0; j < (int)v[k].size(); j++)
            {
                if(pre[v[k][j]] != 0)
                {
                    l[i] = max(l[i], pre[v[k][j]] + 1);
                }
            }
            //if(a[i] == a[i-1]) l[i] = i;
            pre[a[i]] = i;
        }
        clc(pre, 0);
        for(i = n; i > 0; i--)
        {
            r[i] = n;
            k = a[i];
            for(j = 0; j < (int)v[k].size(); j++)
            {
                if(pre[v[k][j]] != 0)
                {
                    r[i] = min(r[i], pre[v[k][j]] - 1);
                }
            }
            //if(a[i]==a[i+1]) r[i] = i;
            pre[a[i]] = i;
        }
        ans = 0;
        for(i = 1; i <= n; i++)
        {
            ans = (ans + (long long)( i - l[i] + 1) * (long long)(r[i] - i + 1) % mod) % mod;
            if(ans > mod) ans -= mod;
        }
        printf("%lld\n", ans % mod);
    }
    return 0;
}

标程代码：
1232ms

#include<cstdio>
#include<iostream>
#include<vector>
#define N 100010
#define P 1000000007
using namespace std;
int n,i,a[N],j,t,q[N];
int l[N],tmp,r[N];
long long a1,a2,ans;
vector<int> vec[10010];
int main()
{
       // freopen("a.in","r",stdin);
  //  freopen("a.out","w",stdout);
    while (scanf("%d",&n)!=EOF)
    {
    for (i=101;i<=10000;i++)
    vec[i].clear();
    for (i=1;i<=n;i++)
    {
        l[i]=0;
        r[i]=n+1;
        scanf("%d",&a[i]);
        if (a[i]>100)
        vec[a[i]].push_back(i);
    }
    for (j=1;j<=100;j++)
    {
        tmp=0;
        for (i=1;i<=n;i++)
        {
            if (a[i]%j==0) l[i]=max(l[i],tmp);
            if (a[i]==j)
            tmp=i;
        }
        tmp=n+1;
        for (i=n;i>=1;i--)
        {
            if (a[i]%j==0) r[i]=min(r[i],tmp);
            if (a[i]==j)
            tmp=i;
        }
    }
    for (i=101;i<=10000;i++)
    q[i]=0;
    for (i=1;i<=n;i++)
    if (a[i]>100)
    {
        for (j=a[i];j<=10000;j=j+a[i])
        while ((q[j]<vec[j].size())&&(vec[j][q[j]]<i))
        {
              r[vec[j][q[j]]]=min(r[vec[j][q[j]]],i);
              if ((q[j]<vec[j].size()-1)&&(vec[j][q[j]+1]<i))
              q[j]++;
              else
              break;
        }
    }
    for (i=101;i<=10000;i++)
    q[i]=vec[i].size()-1;
    for (i=n;i>=1;i--)
    if (a[i]>100)
    {
        for (j=a[i];j<=10000;j=j+a[i])
        while ((q[j]>=0)&&(vec[j][q[j]]>i))
        {
              l[vec[j][q[j]]]=max(l[vec[j][q[j]]],i);
              if ((q[j]>0)&&(vec[j][q[j]-1]>i))
              q[j]--;
              else
              break;
        }
    }
    ans=0;
    for (i=1;i<=n;i++)
    {
        a1=r[i]-i;
        a2=i-l[i];
        ans=(ans+a1*a2)%P;
    }
   printf("%I64d\n",ans);
    }   
}