最大字段和

 

//常州大学新生寒假训练会试
/*

题目描述 
常州大学组织了新生寒假训练一共N天，每天训练可以获得的训练效果是Ei。但是如果连续训练超过K天，萌新们会受不了而被劝退。
现在负责人想知道，如何安排能保证萌新不会被劝退并且能获得最大的训练效果。
输入描述:
第一行：两个用空格隔开的整数：N和K，1≤N≤100000，1≤K≤N
第二行到N+1行：第i+1行有一个整数，表示第N天的训练效果是Ei，(0 <= Ei <= 1,000,000,000）
输出描述:
第一行：单个整数，表示最大的能力之和
示例1
输入
复制
5 2 
1
2
3
4 
5
输出
复制
12
说明
（除了第三天以外每天都在训练，总训练效果为1+2+4+5=12）
备注:
1≤n≤100,000
*/


#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
using namespace std;
#define  ll long long 
#define  N 100009
#define  gep(i,a,b)   for(int  i=a;i<=b;i++)
#define  gepp(i,a,b)  for(int  i=a;i>=b;i--)
#define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
#define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
#define  mem(a,b)  memset(a,b,sizeof(a))
int  n,k,st,ed;
ll a[N],q[N],dp[N];
//dp[i]  :第i天不训练带来的最小损失(逆向思维)
int  main()
{
    scanf("%d%d",&n,&k);
    ll sum=0;
    gep(i,1,n)
    {
        scanf("%lld",&a[i]);
        sum+=a[i];
    }
    //维护一个单调队列
    gep(i,1,n)
    {
        
        while(st<=ed&&dp[q[ed]]>dp[i-1])  ed--;      
        while(q[st]<i-k-1) st++;
        q[++ed]=i-1;
        dp[i]=dp[q[st]]+a[i];//队首最小
    }
    ll ans=0;
    //例如  dp[4]=a[4]+a[1],dp[5]=a[5]+a[2]
    gep(i,n-k,n){//必须从n-k开始，才能保证天数不超过k天
        ans=max(ans,sum-dp[i]);
    }
    printf("%lld
",ans);
    return  0;
}

/ Joy OI / 题目列表 /

最大子序和

题目限制

时间限制	内存限制	评测方式	题目来源
1000ms	131072KiB	标准比较器	Local

题目描述

输入一个长度为n的整数序列，从中找出一段不超过M的连续子序列，使得整个序列的和最大。

例如 1，-3,5,1，-2,3

当m=4时，S=5+1-2+3=7
当m=2或m=3时，S=5+1=6

输入格式

第一行两个数n,m
第二行有n个数，要求在n个数找到最大子序和

输出格式

一个数，数出他们的最大子序和

提示

数据范围：
100%满足n,m<=300000

样例数据

输入样例 #1	输出样例 #1
6 4 1 -3 5 1 -2 3	7

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
using namespace std;
#define  ll long long 
#define  N 300009
const ll inf=9e18;
#define  gep(i,a,b)   for(int  i=a;i<=b;i++)
#define  gepp(i,a,b)  for(int  i=a;i>=b;i--)
#define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
#define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
#define  mem(a,b)  memset(a,b,sizeof(a))
int  n,k,st,ed;
ll a[N],q[N],dp[N],sum[N];
int  main()
{
    scanf("%d%d",&n,&k);
    gep(i,1,n)
    {
        scanf("%lld",&a[i]);
        sum[i]=sum[i-1]+a[i];
    }
    //维护一个单调队列
    gep(i,1,n)
    {       
        while(st<=ed&&sum[q[ed]]>sum[i-1])  ed--;
        while(st<=ed&&q[st]<i-k) st++;//为i服务的
        q[++ed]=i-1;//要非空区间
        dp[i]=sum[i]-sum[q[st]];//队首最小
    }
    ll ans=-inf;
    gep(i,1,n){
        ans=max(ans,dp[i]);
    }
    printf("%lld
",ans);
    return  0;
}

/HDU  3415

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9408    Accepted Submission(s): 3472

Problem Description

Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1].
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

 

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. 
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the minimum length of them.

 

Sample Input

4 6 3 6 -1 2 -6 5 -5 6 4 6 -1 2 -6 5 -5 6 3 -1 2 -6 5 -5 6 6 6 -1 -1 -1 -1 -1 -1

 

Sample Output

7 1 3 7 1 3 7 6 2 -1 1 1

 

Author

shǎ崽@HDU

 

Source

HDOJ Monthly Contest – 2010.06.05

 

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
#include <set>
#include <queue>
using namespace std;
#define  ll long long 
#define  N 200009
#define  gep(i,a,b)   for(int  i=a;i<=b;i++)
#define  gepp(i,a,b)  for(int  i=a;i>=b;i--)
#define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
#define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
#define  mem(a,b)  memset(a,b,sizeof(a))
#define  P  pair<int,int>u+
const ll inf=9e18;
int  n,k,st,ed;
ll a[N],q[N],sum[N];
int t;
int  main()
{
    scanf("%d",&t);
    while(t--){
    scanf("%d%d",&n,&k);
    sum[0]=0;//不用mem(sum,0)
    gep(i,1,n)
    {
        scanf("%lld",&a[i]);
        sum[i]=sum[i-1]+a[i];
    }   
    gep(i,n+1,n+k-1)//环状到n+k-1就可以了
    {
        sum[i]=sum[i-1]+a[i-n];
    }
    //不按照上面求前缀和会超时
    ll ans=-inf;
    ll l,r;
    st=0;
    ed=0;
    mem(q,0);
    //上面三行代码不能少
    gep1(i,1,n+k-1)
    {        
        
        while(st<=ed&&sum[q[ed]]>sum[i-1])  ed--;
        while(st<=ed&&q[st]<i-k) st++;    
        q[++ed]=i-1; 
        if(ans<sum[i]-sum[q[st]]){
        ans=sum[i]-sum[q[st]];
        l=q[st]+1;
        r=i>n?i%n:i;//r可能大于n
        //当出现dp[i]都是最大值时，一定是i小的符合条件
        // 0 2 3 0                         
        }        
    }
    printf("%lld %lld %lld
",ans,l,r);
    }
    return  0;
}

//HDU    6444

Neko's loop

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1420    Accepted Submission(s): 328

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

 

Source

2018中国大学生程序设计竞赛 - 网络选拔赛

 

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <utility>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#define max(x,y) x>=y?x:y
#define lowbit(x) x&(-x)
#define ll long long 
#define ph  push_back
#define   N 100007//因为后面有3*n ,那么最好N 要大与3*n
const ll inf =9e18;
int t;
ll n,m,s,k,a[N],q[N],sum[N];
vector<ll>ve[N];
bool vis[N];
ll solve(int x,int n,ll m){//m  : ll
ll ans=0;
// ll q[N],sum[N]：是错的，因为局部变量必须初始化
for(int i=0;i<=3*n;i++){
    sum[i]=0;
    q[i]=0;
}
for(int i=1;i<=n;i++){
    sum[i]=sum[i+n]=sum[i+2*n]=ve[x][i-1];
}    
for(int i=1;i<=3*n;i++) sum[i]=sum[i-1]+sum[i];//要加到3*n
int st=0,ed=0;
for(int i=1;i<=3*n;i++)//n+ans2(ans2<=2*n)<=3*n
{
    while(st<=ed&&sum[i-1]<sum[q[ed]]) ed--;
    while(st<=ed&&i-q[st]>m) st++;//i-q[st]>m 
    q[++ed]=i-1;//加的是i-1
    ans=max(ans,sum[i]-sum[q[st]]);
}
return ans;
}

int main(){
    scanf("%d",&t);
    for(int tt=1;tt<=t;tt++)
    {
        scanf("%lld%lld%lld%lld",&n,&s,&m,&k);
        for(int i=0;i<N;i++) {
            vis[i]=0;
            ve[i].clear();//每次都要清空
        }
        for(int i=0;i<n;i++){
            scanf("%lld",&a[i]);
        }
        int cnt=0;
        for(int i=0;i<n;i++)
        {
            if(!vis[i])//不然cnt 会不断加
            {
                for(int j=i;!vis[j];j=(j+k)%n)//只要在遇到vis[j]==1就该结束循环了
                {
                    vis[j]=1;
                    ve[cnt].ph(a[j]);
                }
                cnt++;                
            }
        }
        ll ans=-inf;
        for(int i=0;i<cnt;i++)
        {
            ll tmp=0;
            int l=ve[i].size();
            for(int j=0;j<l;j++) tmp+=ve[i][j];//刚开始int i
            ll res=solve(i,l,m);//如果tmp<0，那么跑个不大于m的……即可
            ans=max(ans,res);    
            if(tmp<0) continue;
            ll ans1=m/l;
            ll ans2=m%l;
            if(ans1>=1) {//只要后面还有循环节，且>0就可以全取了
                ans2+=l;//那么最后一个循环要特判
            }
            /*
            例如 ：
            5 1000 5 1
            1 1 -1 -1 1
            正确的为 997，不是999
            */
            tmp=tmp*(max(ans1-1,0ll));//ans1==1可能
            res=max(res,tmp+solve(i,l,ans2));
            ans=max(ans,res);
        }
        ans=max(0ll,s-ans);
        printf("Case #%d: %lld
",tt,ans);
    }
    return 0;
}