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\n",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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <deque>
 8 using namespace std;
 9 #define  ll long long 
10 #define  N 300009
11 const ll inf=9e18;
12 #define  gep(i,a,b)   for(int  i=a;i<=b;i++)
13 #define  gepp(i,a,b)  for(int  i=a;i>=b;i--)
14 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
15 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
16 #define  mem(a,b)  memset(a,b,sizeof(a))
17 int  n,k,st,ed;
18 ll a[N],q[N],dp[N],sum[N];
19 int  main()
20 {
21     scanf("%d%d",&n,&k);
22     gep(i,1,n)
23     {
24         scanf("%lld",&a[i]);
25         sum[i]=sum[i-1]+a[i];
26     }
27     //维护一个单调队列
28     gep(i,1,n)
29     {       
30         while(st<=ed&&sum[q[ed]]>sum[i-1])  ed--;
31         while(st<=ed&&q[st]<i-k) st++;//为i服务的
32         q[++ed]=i-1;//要非空区间
33         dp[i]=sum[i]-sum[q[st]];//队首最小
34     }
35     ll ans=-inf;
36     gep(i,1,n){
37         ans=max(ans,dp[i]);
38     }
39     printf("%lld\n",ans);
40     return  0;
41 }

/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

Recommend

lcy | We have carefully selected several similar problems for you: 3423 3417 3418 3419 3421

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <deque>
 8 #include <set>
 9 #include <queue>
10 using namespace std;
11 #define  ll long long 
12 #define  N 200009
13 #define  gep(i,a,b)   for(int  i=a;i<=b;i++)
14 #define  gepp(i,a,b)  for(int  i=a;i>=b;i--)
15 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
16 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
17 #define  mem(a,b)  memset(a,b,sizeof(a))
18 #define  P  pair<int,int>u+
19 const ll inf=9e18;
20 int  n,k,st,ed;
21 ll a[N],q[N],sum[N];
22 int t;
23 int  main()
24 {
25     scanf("%d",&t);
26     while(t--){
27     scanf("%d%d",&n,&k);
28     sum[0]=0;//不用mem(sum,0)
29     gep(i,1,n)
30     {
31         scanf("%lld",&a[i]);
32         sum[i]=sum[i-1]+a[i];
33     }   
34     gep(i,n+1,n+k-1)//环状到n+k-1就可以了
35     {
36         sum[i]=sum[i-1]+a[i-n];
37     }
38     //不按照上面求前缀和会超时
39     ll ans=-inf;
40     ll l,r;
41     st=0;
42     ed=0;
43     mem(q,0);
44     //上面三行代码不能少
45     gep1(i,1,n+k-1)
46     {        
47         
48         while(st<=ed&&sum[q[ed]]>sum[i-1])  ed--;
49         while(st<=ed&&q[st]<i-k) st++;    
50         q[++ed]=i-1; 
51         if(ans<sum[i]-sum[q[st]]){
52         ans=sum[i]-sum[q[st]];
53         l=q[st]+1;
54         r=i>n?i%n:i;//r可能大于n
55         //当出现dp[i]都是最大值时，一定是i小的符合条件
56         // 0 2 3 0                         
57         }        
58     }
59     printf("%lld %lld %lld\n",ans,l,r);
60     }
61     return  0;
62 }

//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

Input

The first line contains only one integer

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