题目描述：
Neko has a loop of size n.
The loop has a happy value ai on the i−th(0≤i≤n−1) grid.
Neko likes to jump on the loop.She can start at anywhere. If she stands at i−th grid, she will get ai happy value, and she can spend one unit energy to go to ((i+k)modn)−th grid. If she has already visited this grid, she can get happy value again. Neko can choose jump to next grid if she has energy or end at anywhere.
Neko has m unit energies and she wants to achieve at least s happy value.
How much happy value does she need at least before she jumps so that she can get at least s happy value? Please note that the happy value which neko has is a non-negative number initially, but it can become negative number when jumping
输入：
The first line contains only one integer T(T≤50), which indicates the number of test cases.
For each test case, the first line contains four integers n,s,m,k(1≤n≤104,1≤s≤1018,1≤m≤109,1≤k≤n).
The next line contains n integers, the i−th integer is ai−1(−109≤ai−1≤109)
该题关键在于：一个环，环上每点有一个权值，可以从任意点开始走，问至多走m步情况下所经过点的权值和最大是多少。主要比较难想的在于O(nlogn)地求一个环上至多长为k的字段和的最大值。
主要思路：可以在扫一遍的过程中一个个地求以i为结尾的至多长为k的字段和的最大值。维护一个单调队列，保证队头一直是点{i-k..i-k+1....i-i}的前缀和中的最小值，i的前缀和减去该最小值即为i为结尾的至多长为k的字段和的最大值。
单调队列的维护：1.若队头的时间戳与i距离大于k，则队头出队。2.每次将i的前缀和入队的时候，循环进行操作：队尾若大于i的前缀和则队尾出队。
#include<bits/stdc++.h>
#define rep(i,a,b) for(long long i=a;i<=b;++i)
using namespace std;
long long n,m,s,k;
const long long MAXN=1e4+;
long long a[MAXN];
void Input()
{
scanf("%lld%lld%lld%lld",&n,&s,&m,&k);
rep(i,,n-) scanf("%lld",a+i);
}
long long gcd(long long a,long long b)
{
return b==?a:gcd(b,a%b);
}
long long ring[MAXN];
long long head;
long long vis[MAXN];
long long pre[MAXN];
void init()
{
memset(vis,,sizeof(vis));
}
int tail=;
struct Qu
{
long long t,val;
Qu() {}
Qu(long long _val,long long _t)
{
t=_t;
val=_val;
}
}q[*MAXN];
void update(long long idx)
{
Qu one(pre[idx],idx+);
while(tail>=head)
{
if(q[tail].val>=one.val) tail--;
else break;
}
q[++tail]=one;
}
void debug2(long long i)
{
printf("i:%lld\n",i);
rep(i,head,tail)
{
printf("val:%lld t:%lld\n",q[i].val,q[i].t);
}
}
long long calc(long long lim,long long len)
{
// printf("lim:%lld\n",lim);
long long presum=;
rep(i,len,*len-) ring[i]=ring[i-len];
rep(i,,*len-)
{
presum+=ring[i];
pre[i]=presum;
// printf("%lld ",pre[i]);
}
head=;
tail=;
long long maxans=;
q[head]=Qu(,);
rep(i,,*len-)
{
update(i); int nowlen=(i+)-q[head].t;
if(nowlen>lim)
{
head++;
}
// debug2(i); //printf("maxans:%lld val:%lld\n",maxans,pre[i]-q[head].val);
maxans=max(maxans,pre[i]-q[head].val);
}
return maxans; }
void debug(long long len)
{
rep(i,,len-) printf("%d ",ring[i]);
printf("\n");
}
void work(long long icase)
{
long long num=gcd(n,k);
long long ans=;
rep(i,,n-)
{
if(vis[i]) continue;
head=;
long long sum=;
long long len=;
for(long long j=i;!vis[j];j=(j+k)%n)
{
// printf("j:%lld\n",j);
ring[head++]=a[j];
sum+=a[j];
len++;
vis[j]=;
}
long long time=m/len;
long long maxseq;
if(sum<)
{
sum=;
maxseq=calc(min(len - , m), len);
}
else maxseq=calc(m%len,len);
// debug(len);
long long other=calc(len-,len);
if(m>=len)
{
if (other > sum + maxseq){
long long res = (time-) * sum;
long long ians = res + other;
ans = max(ians,ans);
}else{
long long res=time*sum;
long long ians=res+maxseq;
ans=max(ians,ans);
}
}else{
long long res=time*sum;
long long ians=res+maxseq;
ans=max(ians,ans);
}
}
long long finalans=s-ans;
if(finalans<) finalans=;
printf("Case #%lld: %lld\n",icase,finalans);
}
int main()
{
long long T;
scanf("%lld",&T);
rep(tt,,T)
{
init();
Input();
work(tt);
}
return ;
}