2015 Multi-University Training Contest 1 OO’s Sequence
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OO’s Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 3612 Accepted Submission(s): 1330
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 ai mod aj=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 ai(0<ai<=10000)
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)
Output
For each tests: ouput a line contain a number ans.
Sample Input
51 2 3 4 5
Sample Output
23
【题意】给一个序列,求共有多少个找不到任意两个不同数是整除关系的连续子序列,结果mod 1e9+7
【解题方法】维护一下每一个a[i]向左和向右可以到达的最大距离l[i],r[i],那么答案就是对于每一个i,对(i-l[i])*(r[i]-i)求和就可以了。
【AC 代码】
////Created by just_sort 2016/9/12 16:50//Copyright (c) 2016 just_sort.All Rights Reserved//#include <set>#include <map>#include <queue>#include <stack>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 1e5+7;const LL mod = 1e9+7;int a[maxn],vis[maxn],l[maxn],r[maxn];int main(){ int n; while(scanf("%d",&n)!=EOF) { memset(vis,0,sizeof(vis)); int maxx = 0; for(int i=1; i<=n; i++) l[i]=0,r[i]=n+1; for(int i=1; i<=n; i++) scanf("%d",&a[i]); for(int i=1; i<=n; i++){ maxx = max(maxx,a[i]); for(int j=a[i]; j<=maxx; j+=a[i]){ if(vis[j] && r[vis[j]]>i) r[vis[j]] = i; } vis[a[i]] = i; } maxx = 0; memset(vis,0,sizeof(vis)); for(int i=n; i>=1; i--){ maxx = max(maxx,a[i]); for(int j=a[i]; j<=maxx; j+=a[i]){ if(vis[j] && l[vis[j]]<i) l[vis[j]] = i; } vis[a[i]] = i; }// for(int i=1; i<=n; i++){// cout<<l[i]<<" "<<r[i]<<endl;// } LL ans = 0; for(int i=1; i<=n; i++){ ans = (ans+1LL*(i-l[i])*(r[i]-i)%mod)%mod; } printf("%I64d\n",ans); } return 0;}
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