Neko's loop
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1386 Accepted Submission(s): 316
Problem Description
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.
Input
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)
Output
For 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
思路
- 首先通过观察可以发现从i开始走，每次走到（i+k）%n，总可以走回i，存在循环，因此暴力把每个循环处理出来，每处理出来一个循环就计算一个
大佬们都用裴蜀定理定理直接知道了不同循环的数量gcd(n,k),还有每个循环的长度n/gcd(n,k)...orz，看了半天并不知道为什么，求教呀 - 然后就是对每个循环串求最长子段和
- 首先将前缀和处理出来,线段树维护区间最小值（前缀和），每次求i-m～i-1前缀和的最小值，然后用sum[i]-sum[min]更新答案
- 依旧将前缀和处理出来，用递增的单调队列，维护前面i-m~i-1之间前缀和的最小值
- 关于对于循环串的处理
假设循环串的长度为len，能走m步
- 考虑m/len（默认作为横跨两段的串）和m%len两段，我们只需要在一个长度为2*len的循环串中，找到长度不超过m%len的最长连续子段即可，然后看看一整串的和是否大于零，若大于0，则加上m/len个sum[len]（整串的贡献）即可
- 若m>len,则有另外一种可能，就是m%len+m／len个整段这种分配方式并不是最优的，原因在于
假设我们单独取m%len（横跨串）和一整段len来看，可能会存在一个长度大于m%len的横跨串的贡献比m%len+len还大，因此我们可以舍弃一段len，然后使得横跨串的长度从m%len变成len
#include<bits/stdc++.h>
#define ll long long
#define M 10005
using namespace std;
int T,ks,n,tot,m,vi[M],tg[M],len,i,j,k,h,t,Q[M],tp;
ll a[M],s[M<<1],S,ans,re;
int main(){
scanf("%d",&T);
for(ks=1;ks<=T;ks++){
re=-1e16;
memset(vi,0,sizeof(vi));
scanf("%d%lld%d%d",&n,&S,&m,&k);
for(i=0;i<n;i++)scanf("%lld",&a[i]);
for(i=0;i<n;i++){
len=0;
for(j=i;;j=(j+k)%n){
if(!vi[j]){
vi[j]=1;
tg[++len]=j;
}
else break;
}
if(len==0)continue;
for(j=1;j<=2*len;j++){
if(j<=len)s[j]=s[j-1]+a[tg[j]];
else s[j]=s[j-1]+a[tg[j-len]];
}
ans=0;
h=t=0;Q[t++]=0;tp=m;
if(s[len]>=0){ans+=s[len]*(tp/len);tp%=len;}
else tp=min(tp,len);
for(j=1;j<=2*len;j++){
while(h<t&&j-tp>Q[h])h++;
if(h<t)re=max(re,ans+s[j]-s[Q[h]]);
while(h<t&&s[Q[t-1]]>=s[j])t--;
Q[t++]=j;
}
if(m>=len){
tp=m-len;ans=0;
if(s[len]>=0)ans+=s[len]*(tp/len);
tp=len;
h=t=0;Q[t++]=0;
for(j=1;j<=2*len;j++){
while(h<t&&j-tp>Q[h])h++;
if(h<t)re=max(re,ans+s[j]-s[Q[h]]);
while(h<t&&s[Q[t-1]]>=s[j])t--;
Q[t++]=j;
}
}
}
printf("Case #%d: %lld
",ks,max(0ll,S-re));
}
}