题目链接:
题目大意:
给出一个数组,f(l,r)表示区间中i的数量,i满足条件,a[i]%a[j]!=0(j!=i , l<=j<=r)
∑i=1n∑j=inf(i,j) mod (109+7).
题目分析:
首先就是枚举每个i,然后算取每个i对于最后结果的贡献,也就是找到i坐标之前a[i]%a[l]的最右的位置x,i坐标之后a[i]%a[j]的最左位置y,那么对于区间的左边界为[l,i],右边界为[i,r]的区间都有贡献,所以对于最后答案的贡献就是(i-x+1)*(j-y+1)
那么找位置的操作如何做?
就是从前到后扫描数组,利用mark[]数组记录某个值出现的最右位置,然后每次枚举i的约数,然后利用Mark数组更新当前的最右边界即可
代码如下:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define MAX 100007
using namespace std;
typedef long long LL;
int mark[MAX];
int a[MAX];
LL l[MAX];
LL r[MAX];
int n;
const LL mod = 1e9+7;
LL ans;
int main ( )
{
while ( ~scanf ( "%d" , &n ) )
{
ans = 0;
int loc;
memset ( mark , -1 , sizeof(mark));
for ( int i = 0 ; i < n ; i++ )
scanf ( "%d" , &a[i] );
for ( int i = 0 ; i < n ; i++ )
{
loc = -1;
for ( int j = 1 ; j*j <= a[i] ; j++ )
{
if ( a[i]%j ) continue;
if (mark[j] != -1 )
loc = max ( loc, mark[j] );
int x = a[i]/j;
if ( mark[x] != -1 )
loc = max ( loc , mark[x] );
}
l[i] = loc;
mark[a[i]] = i;
}
memset ( mark , -1 , sizeof ( mark ));
for ( int i = n-1 ; i >= 0 ; i-- )
{
loc = n;
for ( int j = 1 ; j*j <= a[i] ; j++ )
{
if ( a[i]%j ) continue;
if ( mark[j] != -1 )
loc = min ( loc , mark[j] );
int x = a[i]/j;
if ( mark[x] != -1 )
loc = min ( loc , mark[x] );
}
r[i] = loc;
mark[a[i]] = i;
}
for ( LL i = 0 ; i < n ; i++ )
{
ans += (i-l[i])*(r[i]-i)%mod;
ans %= mod;
}
printf ( "%I64d\n" , ans );
}
}