HDU 5288——OO’s Sequence——————【技巧题】

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1549 Accepted Submission(s): 559


[align=left]Problem Description[/align]
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=1n∑j=inf(i,j) mod （109+7）.

[align=left]Input[/align]
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)

[align=left]Output[/align]
For each tests: ouput a line contain a number ans.

[align=left]Sample Input[/align]

5

1 2 3 4 5

[align=left]Sample Output[/align]

23

题目大意：给你一个长度为n的区间。求这个区间内任意子区间中ai没有因子的ai的个数。

解题思路：从左往右遍历a数组得到L数组。对于ai，枚举j 1-->sqrt（ai），如果j是ai的约数并且j已经在ai之前，那么我用max取离i最近的位置；相应的ai/j也是ai的约数，同理如果ai/j也在ai之前，也用max取离i最近的位置。对于R数组，同理可得。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
const int maxn=1e5+20;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
int a[maxn],L[maxn],R[maxn];
int pos[maxn];
int main(){
int n,limj;
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
memset(pos,0,sizeof(pos));
for(int i=1;i<=n;i++){
L[i]=0;
limj=(int)sqrt(a[i]);
for(int j=1;j<=limj;j++){
if(a[i]%j==0){
if(pos[j]){
L[i]=max(L[i],pos[j]);
}
if(pos[a[i]/j]){
L[i]=max(L[i],pos[a[i]/j]);
}
}
}
pos[a[i]]=i;
}
memset(pos,0,sizeof(pos));
for(int i=n;i>=1;i--){
R[i]=n+1;
limj=(int)sqrt(a[i]);
for(int j=1;j<=limj;j++){
if(a[i]%j==0){
if(pos[j]){
R[i]=min(R[i],pos[j]);
}
if(pos[a[i]/j]){
R[i]=min(R[i],pos[a[i]/j]);
}
}
}
pos[a[i]]=i;
}
int sum=0,res;
for(int i=1;i<=n;i++){
sum=(sum%MOD+((i-L[i])*(R[i]-i))%MOD)%MOD;
}
printf("%d\n",sum);
}
return 0;
}