Codeforces Round #464 (Div. 2) C. Convenient For Everybody

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.InputThe 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).OutputOutput 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.ExamplesinputCopy

3
1 2 3
1 3

output

3

inputCopy

5
1 2 3 4 1
1 3

output

4

NoteIn 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 and 2hours 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.
题意：在未来，一天一共有n个小时，也一共有n个时区，每两个相邻的时区之间差是一个小时。现在有一些网站想要举办在线编程比赛，希望有最多的人能参与。因为人们以自己的时区的时间为标准，只参加在s点以后（包括s点）开始的且在f点之前（包括f点）的比赛。每场比赛的时间为1个小时，让你求出比赛在什么时间开始会有最多的人参加，输出以第一个时区的时间为标准的开始时间。
思路：因为比赛时间为1个小时，所以显然能够参加到比赛的时区一共有 (f - s) 个。这样肯定就是求连续(f-s)个ai的最大是从哪个开始的。这题有一个坑点，就是第二天的1点是算是比当天的2点要早的，第二天的2点是比当天的1点要迟的，所以算连续的时候，如果后面不够了，要补上前面的。还有就是要输出最小的那个答案。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;

LL sum[100010];

int main(void)
{
int n,i,j;
while(scanf("%d",&n)==1)
{
sum[0] = 0;
for(i=1;i<=n;i++)
{
int x;
scanf("%d",&x);
sum[i] = sum[i-1] + x;
}
int s,f;
scanf("%d%d",&s,&f);
int t = f-s;
LL maxn = 0;
int ans = n;
for(i=1;i<=n-t+1;i++)
{
LL tmp = sum[i+t-1] - sum[i-1];
if(tmp > maxn)
{
maxn = tmp;
ans = s - i + 1;
if(ans <= 0)
ans += n;
}
else if(tmp == maxn)
{
int ta = s-i+1;
if(ta <= 0)
ta += n;
if(ta < ans)
ans = ta;
}
}
for(;i<=n;i++)
{
LL tmp = sum
- sum[i-1] + sum[t-n+i-1];
if(tmp > maxn)
{
maxn = tmp;
ans = s - i + 1;
if(ans <= 0)
ans += n;
}
else if(tmp == maxn)
{
int ta = s-i+1;
if(ta <= 0)
ta += n;
if(ta < ans)
ans = ta;
}
}

printf("%d\n",ans);
}

return 0;
}