Codeforces Round #209 (Div. 2) D. Pair of Numbers (模拟)
2 seconds
256 megabytes
standard input
standard output
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
- there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
- value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an(1 ≤ ai ≤ 106).
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
54 6 9 3 6
1 32
51 3 5 7 9
1 41
52 3 5 7 11
5 01 2 3 4 5
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).
【分析】给定一个正整数数组,求最长的区间,使得该区间内存在一个元素,它能整除该区间的每个元素。
可能我的方法有点复杂吧。我是先从小到大排序,记录所有数的位置,然后从小到大向两边扩展,就行了,详细见代码一。还有种做法类似DP,很短很神奇,见代码二。
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <queue> #include <vector> #define inf 0x3f3f3f3f #define met(a,b) memset(a,b,sizeof a) #define pb push_back #define mp make_pair typedef long long ll; using namespace std; const int N = 3e5+10; const int M = 1e6+10; int n,m,k,tot=0,tim=0; int head[N],vis[N],l[N],r[N]; int a[N],b[N]; vector<pair<int,int> >vec; vector<int>Ans; struct man{int l,r,len; }ans[N]; bool cmp(man s,man d){return s.len>d.len; } void Find(int p) {vis[p]=1;int ret=0;int u=b[p];int ll,rr;++tot;for(int i=1; i<=n; i++) {if(!l[tot]) {ll=p-i;if(ll>=1&&b[ll]%u==0) {vis[ll]=1;} else l[tot]=ll+1,ret++;}if(!r[tot]) {rr=p+i;if(rr<=n&&b[rr]%u==0) {vis[rr]=1;} else r[tot]=rr-1,ret++;}if(ret==2)return;} } int main() {scanf("%d",&n);for(int i=1; i<=n; i++) {scanf("%d",&a[i]);b[i]=a[i];vec.pb(mp(a[i],i));}sort(vec.begin(),vec.end());for(int i=0;i<n;i++){int p=vec[i].second;if(!vis[p])Find(p);}for(int i=1; i<=tot; i++) {ans[i-1].l=l[i];ans[i-1].r=r[i];ans[i-1].len=r[i]-l[i];}sort(ans,ans+tot,cmp);int t=ans[0].len,ret=0;for(int i=0;i<tot;i++){if(ans[i].len<t)break;ret++;}printf("%d %d\n",ret,t);for(int i=0;i<ret;i++){Ans.pb(ans[i].l);}sort(Ans.begin(),Ans.end());for(int i=0;i<ret;i++){printf("%d ",Ans[i]);}printf("\n");return 0; }
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <queue> #include <vector> #define inf 0x3f3f3f3f #define met(a,b) memset(a,b,sizeof a) #define pb push_back typedef long long ll; using namespace std; const int N = 2e6+10; const int M = 1e6+10; int n,L,R,len,Ans,xllend3; int a[N],ans_l[N],ans_r[N],ans[N]; void init() {scanf("%d",&n);for(int i=1; i<=n; ++i)scanf("%d",&a[i]),ans_l[i]=ans_r[i]=i; } void work() {for(int i=1; i<=n; ++i)for(; ans_l[i]>1&&a[ans_l[i]-1]%a[i]==0;)ans_l[i]=ans_l[ans_l[i]-1];for(int i=n; i>=1; --i)for(; ans_r[i]<n&&a[ans_r[i]+1]%a[i]==0;)ans_r[i]=ans_r[ans_r[i]+1];for(int i=1; i<=n; ++i) {L=ans_l[i];R=ans_r[i];if(R-L==len)ans[++Ans]=L;if(R-L>len)len=R-L,ans[Ans=1]=L;} }void outit() {sort(ans+1,ans+Ans+1);for(int i=1; i<=Ans; ++i)if(ans[i]!=ans[i-1])++xllend3;printf("%d %d\n%d",xllend3,len,ans[1]);for(int i=2; i<=Ans; ++i)if(ans[i]!=ans[i-1])printf(" %d",ans[i]);puts(""); }int main() {init();work();outit();return 0; }
转载于:https://www.cnblogs.com/jianrenfang/p/6479174.html
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