2015-year-school joint training first game Oo ' s Sequence (hdu5288)

Test instructions: Given a sequence of length n, which specifies that F (l,r) is for a l,r in the range of a number a[i], can not find a corresponding j to make a[i]%a[j]=0, then l,r within the number of i,f (L,R) is a few. Ask what the sum of all F (l,r) is.
The interval given in the formula is the interval of all existence.

Idea: Direct enumeration of each number, for this number, if the number is legal I, then the maximum length to the left can expand the maximum length is how much, then I is legal case is the left length * right length (including I and I is the number of valid interval).

Statistics left length can be judged a[i] of the approximate whether in the previous appeared ... Because a[i]<=10000, you can tag all the numbers on the left of I with an array a[k],k

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace STD;Const intMax =1e5+ -;Const intMoD =1e9+7;intA[max];intUsed[max];Long LongZuo[max];Long LongYou[max];intCalintx) {intm =sqrt(x*1.0);intMaxi = max (used[1],USED[X]); for(inti =2; I <= m; ++i) {if(x% i = =0{maxi = max (max (Maxi,used[i]), used[x/i]); }    }returnMaxi;}intMain () {intN while(~scanf("%d", &n)) {memset(Used,0,sizeof(used));memset(Zuo,0,sizeof(Zuo));memset(You,0,sizeof(you)); for(inti =1; I <= N; ++i) {scanf("%d", a+i);             Zuo[i] = i-cal (A[i]);        Used[a[i]] = i; }memset(Used,0,sizeof(used));//Turn to the right, n-i+1, you know the benefits of the simulation         for(inti =1; I <= N; ++i) {you[n-i+1] = i-cal (a[n-i+1]); used[a[n-i+1]] = i; }Long LongAns =0; for(inti =1; I <= N;        ++i) {ans + = (zuo[i]*you[i])%mod; }printf("%lld\n", Ans%mod); }return 0;}

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2015-year-school joint training first game Oo ' s Sequence (hdu5288)