Codeforces Round #464 (Div. 2)
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are nplanes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles.
思路：跑一边tarjin，判断其中是否有点数为3个的强连通分量。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 5010 using namespace std; int n,tot,tim,top,ans,sumcol; int low[MAXN],dfn[MAXN],col[MAXN]; int to[MAXN],net[MAXN],head[MAXN]; int stack[MAXN],vis[MAXN],visstack[MAXN]; void add(int u,int v){ to[++tot]=v;net[tot]=head[u];head[u]=tot; } void tarjin(int now){ low[now]=dfn[now]=++tim; stack[++top]=now; vis[now]=1;visstack[now]=1; for(int i=head[now];i;i=net[i]) if(!vis[to[i]]){ tarjin(to[i]); low[now]=min(low[now],low[to[i]]); } else if(visstack[to[i]]) low[now]=min(low[now],dfn[to[i]]); if(low[now]==dfn[now]){ sumcol++; col[now]=sumcol; while(stack[top]!=now){ col[stack[top]]=sumcol; visstack[stack[top]]=0; top--; } top--; visstack[now]=0; } } int main(){ scanf("%d",&n); for(int i=1;i<=n;i++){ int x; scanf("%d",&x); add(i,x); } for(int i=1;i<=n;i++) if(!vis[i]) tarjin(i); memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) vis[col[i]]++; for(int i=1;i<=sumcol;i++) if(vis[i]==3) ans++; if(ans) cout<<"YES"; else cout<<"NO"; }
Dima has a hamsters farm. Soon N hamsters will grow up on it and Dima will sell them in a city nearby.
Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters.
Dima can buy boxes at a factory. The factory produces boxes of K kinds, boxes of the i-th kind can contain in themselves ai hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount.
Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm.
Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.
The first line contains two integers N and K (0 ≤ N ≤ 1018, 1 ≤ K ≤ 105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces.
The second line contains K integers a1, a2, ..., aK (1 ≤ ai ≤ 1018 for all i) — the capacities of boxes.
Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to K in the order they are given in input.
If there are many correct answers, output any of them.
思路：o(k)扫一遍就好了。
错误：minn值取小了，3个变量名重复竟然没看见。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; long long k,ans,bns,n,minn=1223372036854775807; int main(){ cin>>n>>k; for(int i=1;i<=k;i++){ long long x;cin>>x; long long kkk=n%x; if(kkk<minn){ minn=kkk;ans=i;bns=x; } } long long kk=n/bns; cout<<ans<<" "<<kk; }
In distant future on Earth day lasts for n hours and that's why there are n timezones. Local times in adjacent timezones differ by one hour. For describing local time, hours numbers from 1 to n are used, i.e. there is no time "0 hours", instead of it "n hours" is used. When local time in the 1-st timezone is 1 hour, local time in the i-th timezone is i hours.
Some online programming contests platform wants to conduct a contest that lasts for an hour in such a way that its beginning coincides with beginning of some hour (in all time zones). The platform knows, that there are ai people from i-th timezone who want to participate in the contest. Each person will participate if and only if the contest starts no earlier than s hours 00 minutes local time and ends not later than f hours 00 minutes local time. Values s and f are equal for all time zones. If the contest starts at f hours 00 minutes local time, the person won't participate in it.
Help platform select such an hour, that the number of people who will participate in the contest is maximum.
The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of hours in day.
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 10 000), where ai is the number of people in the i-th timezone who want to participate in the contest.
The third line contains two space-separated integers s and f (1 ≤ s < f ≤ n).
Output a single integer — the time of the beginning of the contest (in the first timezone local time), such that the number of participants will be maximum possible. If there are many answers, output the smallest among them.
In the first example, it's optimal to start competition at 3 hours (in first timezone). In this case, it will be 1 hour in the second timezone and2 hours in the third timezone. Only one person from the first timezone won't participate.
In second example only people from the third and the fourth timezones will participate.
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 100010 using namespace std; int a[MAXN]; int n,s,f,ans,sum,maxn; int main(){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); scanf("%d%d",&s,&f); for(int i=1;i<=n;i++){ int x=i; if(x>=s&&x<=f-1) sum+=a[1]; for(int j=1;j<n;j++){ x++;if(x>n) x=1; if(x>=s&&x<=f-1) sum+=a[j+1]; } if(sum>maxn){ ans=i;maxn=sum; }sum=0; } cout<<ans; }