Neko has a loop of size nn.
The loop has a happy value aiai on the i−th(0≤i≤n−1)i−th(0≤i≤n−1) grid.
Neko likes to jump on the loop.She can start at anywhere. If she stands at i−thi−thgrid, she will get aiai happy value, and she can spend one unit energy to go to ((i+k)modn)−th((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 mm unit energies and she wants to achieve at least ss happy value.
How much happy value does she need at least before she jumps so that she can get at least ss 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.
InputThe first line contains only one integer T(T≤50)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)n,s,m,k(1≤n≤104,1≤s≤1018,1≤m≤109,1≤k≤n).
The next line contains nn integers, the i−thi−th integer is ai−1(−109≤ai−1≤109)ai−1(−109≤ai−1≤109)
OutputFor each test case, output one line "Case #x: y", where x is the case number (starting from 1) and y is the answer.Sample Input
2 3 10 5 2 3 2 1 5 20 6 3 2 3 2 1 5
Sample Output
Case #1: 0 Case #2: 2 题意：给n个数字，当位于某一个数字时，可以得到这个数字的快乐值（可以重复获得），可以走m步，每次向后跳k格（循环），问快乐值要达到s，那么初始快乐值的最小值是多少？思路：这是2018ccpc的网络赛的题，暑假看着学长打比赛，在傍边连读题都帮不上。。。。 首先有一个显而易见的结论，对于某一个起点，跳着跳着就会回到起点，也就是说，一定存在循环节。其次显而易见的，就是对于某一个数，只会存在于一个循环节中。那么如果不考虑循环节的起点差异，找出所有循环节就是O(n)的。而实际上我们确实不用考虑，原因接下来再说。 我们可以轻易的算出某个循环节中所有元素的和--sum，因为存在负数，所以sum可能是负数。再声明一下，len是循环节长度。若sum<0，那么最多只需走min（m,len）步就可以得到最优解。若sum>0,则需要先走m/len-1圈，为什么减一，其他博客有详解（博主太懒了），然后最多走m-(m/len-1)*len)步
先说这后面多出来的步数，假设是p步，这p步在sum<0时，p<=len，sum<0时,len<=p<=2*len。为了求出这p步走出的最大值，我们使用单调队列。首先求出循环节的前缀和sum，单调队列维护长度为p的滑窗的sum最小值。维护 ans =（ sum[i]-单调队列最小值 ） 的最大值就可以啦！由于p的长度，要把循环节扩展3倍哟。 然后我们看到，这个单调队列，在ans取得最大值时，单调队列的最小值位置是不定的，所以很容易想到，不一定就是循环节找出来时的起点，所以我们很容易想到，这一个过程，相当于已经枚举了实际问题中的起点。也就是，我们在取ans的最大值时的最小值位置，就可以当实际问题中的起点。
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int maxm = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
ll num[maxn];
vector<ll>vec;
bool vis[maxn];
ll sum[maxn];
ll n,s,m,k;
int len;
struct node
{
ll x;
int id;
};
deque<node>q;
ll solve(int p){
q.clear();
for(int i=0;i<len;i++){
vec.push_back(vec[i]);
}
for(int i=0;i<len;i++){
vec.push_back(vec[i]);
}
ll ans=0;
sum[0]=vec[0];
for(int i=1;i<len*3;i++){
sum[i]=sum[i-1]+vec[i];
}
for(int i=0;i<len*3;i++){
while(!q.empty()&&q.back().x>sum[i]){
q.pop_back();
}
if(!q.empty()&&q.front().id<i-p){q.pop_front();}
q.push_back(node{sum[i],i});
ans=max(ans,1ll*sum[i]-q.front().x);
}
return ans;
}
int main()
{
int T;
scanf("%d",&T);
int cases=0;
while(T--){
ll ans=0;
cases++;
scanf("%lld%lld%lld%lld",&n,&s,&m,&k);
for(int i=1;i<=n;i++){
scanf("%lld",&num[i]);
}
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++){
vec.clear();
if(vis[i]){continue;}
int pos=i;
for(int j=1;j<=m;j++){
vec.push_back(num[pos]);
vis[pos]=true;
pos+=k;
if(pos>n){pos%=n;}
if(vis[pos]){break;}
}
len=vec.size();
ll sum=0;
for(int i=0;i<len;i++){
sum+=vec[i];
}
if(sum<0){
ans=max(ans,solve(len));
}
else{
ans=max(ans,solve(m-(m/len-1)*len)+(m/len-1)*sum);
}
}
printf("Case #%d: %lld\n",cases,max(0ll,s-ans));
}
return 0;
}
HDU 6444 Neko's loop（单调队列）
原文地址：https://www.cnblogs.com/ZGQblogs/p/10806800.html