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OO’s Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 353 Accepted Submission(s): 117
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 a
i mod a
j=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 a
i(0
i<=10000)
Output For each tests: ouput a line contain a number ans.
Sample Input
5
1 2 3 4 5
Sample Output
23
f(l,r)求[l,r]区间内存在多少i（区间内没有i的约数）。公式中给出的区间，也就是所有存在的区间。
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记录l[i]，r[i]，i的约数存在的位置，要求最接近i的。可以用数组记录下出现的最接近当前位置的数。得到l[i],r[i]后，那么i可以被计数的区间也就有了，左边界为l[i]到i，右边界为i到r[i]，那么i一共会被记录(i-l[i])*(r[i]-i)次，累加所有的计数。
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#include
#include
#include
using namespace std ; #define LL __int64 const int MOD = 1e9+7 ; int a[100010] , l[100010] , r[100010] ; int Map[10010] ; int main() { int i , j , n ; LL ans ; while( scanf(%d, &n) != EOF ) { for(i = 1 ; i <= n ; i++) scanf(%d, &a[i]) ; memset(Map,0,sizeof(Map)) ; for(i = 1 ; i <= n ; i++) { for(j = 1 , l[i] = 0 ; j*j <= a[i] ; j++) { if( a[i]%j ) continue ; if( Map[j] ) l[i] = max(l[i],Map[j]) ; if( Map[a[i]/j] ) l[i] = max(l[i],Map[a[i]/j]) ; } Map[a[i]] = i ; } memset(Map,0,sizeof(Map)) ; for(i = n ; i > 0 ; i--) { for(j = 1 , r[i] = n+1 ; j*j <= a[i] ; j++) { if( a[i]%j ) continue ; if( Map[j] ) r[i] = min(r[i],Map[j]) ; if( Map[a[i]/j] ) r[i] = min(r[i],Map[a[i]/j]) ; } Map[a[i]] = i ; } for(i = 1 , ans = 0 ; i <= n ; i++) { ans = (ans + (i-l[i])*(r[i]-i) ) % MOD ; } printf(%I64d , ans) ; } return 0 ; }
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