多校1 OO’s Sequence 1001

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 256 Accepted Submission(s): 94


[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

[align=left]Source[/align]
2015 Multi-University Training Contest 1

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#include <math.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int mod=1e9+7;

int main()
{
int n,i,j,k;
int str[100005],spfa[10005],l[100005],r[100005];
int s;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
scanf("%d",&str[i]);
for(i=1;i<=10000;i++)
spfa[i]=0;
for(i=1;i<=n;i++)
l[i]=0,r[i]=n+1;
for(i=1;i<=n;i++)
{
for(j=1;j<=sqrt(str[i]);j++)
{
if(str[i]%j==0)
{
if(spfa[j]>l[i])
l[i]=spfa[j];
if(spfa[str[i]/j]>l[i])
l[i]=spfa[str[i]/j];
}
}
spfa[str[i]]=i;
}
for(i=10000;i>=1;i--)
spfa[i]=n+1;
for(i=n;i>=1;i--)
{
for(j=1;j<=sqrt(str[i]);j++)
{
if(str[i]%j==0)
{
if(spfa[j]<r[i])
r[i]=spfa[j];
if(spfa[str[i]/j]<r[i])
r[i]=spfa[str[i]/j];
}
}
spfa[str[i]]=i;
}
s=0;
for(i=1;i<=n;i++)
{
s=(s+(i-l[i])*(r[i]-i)%mod)%mod;
}
printf("%d\n",s);
}
return 0;
}

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