OO ' s Sequence

OO ' s Sequence

Topic Abstract: Gives you an integer sequence. For each of its non-empty subset intervals, how many integers are not divisible by the other number. The sum of these numbers for each non-empty set interval.

Analysis: Use L[i],r[i] to represent the nearest approximate in A[i].

Timeout code:

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6#include <string>7#include <vector>8#include <Set>9#include <map>Ten#include <queue> One#include <sstream> A using namespacestd; -typedefLong LongLL; - Const intINF =0x4ffffff; the Const DoubleEXP = 1e-5; - ConstLL mod = 1e9+7; - Const intms= 1e5+5; - Const intMS2 = 1e4+1; +  - intA[ms]; + intL[ms]; A intR[ms]; at  - intMain () - { -       intN; -        while(SCANF ("%d", &n)! =EOF) -       { in              for(inti =1; i<=n;i++) -scanf"%d",&a[i]); toa[0] =1; +a[n+1] =1; -  the              for(inti =1; i<=n;i++) *             { $                    for(intj = I1; j>=0; j--)Panax Notoginseng                   { -                         if(A[i]% a[j] = =0) the                         { +L[i] =J; A                                Break; the                         } +                   } -  $                    for(intj = i+1; j<=n+1; j + +) $                   { -                         if(A[i]% a[j] = =0) -                         { theR[i] =J; -                                Break;Wuyi                         } the                   } -             } WuLL ans=0; -              for(inti =1; i<=n;i++) About             { $Ans + = (i-l[i]) * (R[i]-i); -Ans%=MoD; -             } -printf"%lld\n", ans); A       } +       return 0; the}

Optimization: Since each integer is not greater than 10000, we can preprocess the approximate number of each. In processing l[i] and r[i], use the pre array to represent the integers that appear earlier. See Code.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6#include <string>7#include <vector>8#include <Set>9#include <map>Ten#include <queue> One#include <sstream> A using namespacestd; -typedefLong LongLL; - Const intINF =0x4ffffff; the Const DoubleEXP = 1e-5; - ConstLL mod = 1e9+7; - Const intms= 1e5+5; - Const intMS2 = 1e4+1; +  - intA[ms]; + intL[ms]; A intR[ms]; at intPre[ms]; -vector<int>Vec[ms]; -  - voidInit () - { -        for(inti =1; i<ms2;i++) in       { -              for(intj = I;j <ms2;j+=i) to             { + Vec[j].push_back (i); -             } the       } * } $ Panax Notoginseng intMain () - { the init (); +       intN; A        while(SCANF ("%d", &n)! =EOF) the       { +              for(inti =1; i<=n;i++) -scanf"%d",&a[i]); $memset (PRE,0,sizeof(pre)); $              for(inti =1; i<=n;i++) -             { -                   intt =0; the                    for(intj =0; J<vec[a[i]].size (); j + +) -                   {Wuyi                         if(Pre[vec[a[i]][j]]) thet =Max (T,pre[vec[a[i]][j]]); -                   } WuL[i] =T; -Pre[a[i]] =i; About             } $  -memset (PRE,0,sizeof(pre)); -              for(inti = n;i>0; i--) -             { A                   intt = n+1; +                    for(intj =0; J<vec[a[i]].size (); j + +) the                   { -                         if(Pre[vec[a[i]][j]]) $                         { thet =min (t,pre[vec[a[i]][j]]); the                         } the                   } theR[i] =T; -Pre[a[i]] =i; in             } theLL ans=0; the              for(inti =1; i<=n;i++) About             { theAns + = (i-l[i]) * (R[i]-i); theAns%=MoD; the             } +printf"%lld\n", ans); -       } the       return 0;Bayi}

OO ' s Sequence