HDU6098 Inversion

Inversion

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 45    Accepted Submission(s): 28

Problem Description

Give an array A, the index starts from 1.
Now we want to know Bi=maxi∤jAj , i≥2.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integer n : the size of array A.
Next one line contains n integers, separated by space, ith number is Ai.

Limits
T≤20
2≤n≤100000
1≤Ai≤1000000000
∑n≤700000

 

Output

For each test case output one line contains n-1 integers, separated by space, ith number isBi+1.

 

Sample Input

241 2 3 441 4 2 3

 

Sample Output

3 4 32 4 4

 

题目大意：给你一个数列A下标为j然后i从2开始让你求j%i!=0的A数列中的最大值为Bi，i从2开始到n。

解题思路：我的想法是分区间求最大值将j%i==0的数除去然后这个数列就被分为了好几个区间让后用RMQ求区间的最大值，要注意数组边缘的值。

AC代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;#define MAXN 100010int n,m;int a[MAXN];int b[MAXN];int f[20][MAXN<<1];void ST(){    for(int i=1; i<=n; i++)    {        f[0][i]=a[i];    }    for(int j=1; j<20; j++)    {        for(int i=1; i<=n; i++)        {            f[j][i]=max(f[j-1][i],f[j-1][i+(1<<(j-1))]);        }    }}void init(){    memset(f,0,sizeof(f));    memset(a,0,sizeof(a));}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        init();        memset(b,0,sizeof(b));        for(int i=1; i<=n; i++)        {            scanf("%d",&a[i]);        }        ST();        for(int i=2; i<=n; i++)        {            for(int j=0; j<=n/2; j++)            {                if(i*j>n)                {                    break;                }                else                {                    if(i*j<n&&(j+1)*i>n)                    {                        int l=j*i+1;                        int r=n;                        int k=(int)(log((double)(r-l+1))/log(2.0));                        b[i]=max(b[i],max(f[k][l],f[k][r-(1<<k)+1]));                    }                    else                    {                        int l=j*i+1;                        int r=(j+1)*i-1;                        int k=(int)(log((double)(r-l+1))/log(2.0));                        b[i]=max(b[i],max(f[k][l],f[k][r-(1<<k)+1]));                    }                }            }            //printf("%d\n",b[i]);        }        for(int i=2;i<n;i++){printf("%d ",b[i]);}printf("%d\n",b[n]);    }    return 0;}

还有一种思路为把A数列存到一个结构体数组里面，结构体数组里面分别存的是A数列的值以及A数列的下标。然后给A数列排序，找的第一个下标不能被j整除的数就是Bi的值。

这个应该为正解，比较好想，上一个虽然也能过但感觉就不怎么好了。写上面的思路是因为昨天刚刚看了RMQ算法今天就能用上感觉没有白学。

代码：

#include<cstdio>#include<algorithm>using namespace std;struct ss{    int i;    long long num;};bool cmp(ss a,ss b){    return a.num>b.num;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n;        ss a[100000+5];        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            a[i].i=i;            scanf("%lld",&a[i].num);        }        sort(a+1,a+n+1,cmp);        for(int j=2;j<=n;j++)        {            for(int k=1;k<=n;k++)            {                if(a[k].i%j!=0)                {                    printf("%lld%c",a[k].num,j==n?'\n':' ');                    break;                }            }        }    }    return 0;}