hdu 5288 OO’s Sequence(2015 Multi-University Training Contest 1)
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OO’s Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1080 Accepted Submission(s): 403
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=1n∑j=inf(i,j) mod (109+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)
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
51 2 3 4 5
Sample Output
23
Author
FZUACM
Source
2015 Multi-University Training Contest 1
题目大意:
给出一段序列,它的所有非空且里面的元素也为连续的子集中,不存在它可以整除数的元素的个数的和。
解题思路:
对序列的每个元素分开考虑,对于每个元素,往左边找可以找到连续区间长度为a的序列(包括这个元素),往右边可
以找到长度为b的序列(包括这个元素),那么这个元素出现了a*b次,将每个元素出现的次数相加就可以了。如何找这个区
间?发现序列元素值就10000,于是可以开个10000的数组,数组记录这个值最后一次出现的位置,然后枚举约数就
就可以了。
代码:
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int inf=0x7fffffff;const int maxn=100000+1000;const int mod=1000000000+7;int a[maxn];long long l[maxn];long long r[maxn];int h[maxn];int main(){ int n; while(~scanf("%d",&n)) { memset(h,0,sizeof(h)); for(int i=1;i<=n;i++) scanf("%d",&a[i]); int cur=inf,te; for(int i=1;i<=n;i++) { cur=inf; for(int j=1;j*j<=a[i];j++) { if(a[i]%j==0) { cur=min(cur,i-h[j]); te=a[i]/j; cur=min(cur,i-h[te]); } } l[i]=cur; h[a[i]]=i; } for(int i=1;i<=10500;i++) h[i]=n+1; for(int i=n;i>0;i--) { cur=inf; for(int j=1;j*j<=a[i];j++) { if(a[i]%j==0) { cur=min(cur,h[j]-i); te=a[i]/j; cur=min(cur,h[te]-i); } } h[a[i]]=i; r[i]=cur; } long long ans=0; for(int i=1;i<=n;i++) { ans=(ans+(l[i]*r[i]))%mod; } printf("%I64d\n",ans); } return 0;}
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