HDU 1003 - Max Sum

Posted 2020-06-09

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second 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 first one. Output a blank line between two cases.

 

Sample Input

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1: 14 1 4 Case 2: 7 1 6

 

 

大致题意：

　　给一个序列，找出他的和最大的子序列；

解题思路：

　　以下有两个代码：

　　第一个代码：

　　　　贪心，遍历所有可能，找最大，当然会超时，所以要剪枝；复杂度O（n^2);

　　第二个代码：

　　　　 比较神奇，DP，复杂度0（n）；

　　　　以a[0]结尾的子序列只有a[0]
　　　　以a[1]结尾的子序列有 a[0]a[1]和a[1]
　　　　以a[2]结尾的子序列有 a[0]a[1]a[2] / a[1]a[2] / a[2]

　　　　很容易得出状态转移方程式 sum_a[i]=max(sum_a[i-1]+a[i],a[i]);

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 int main()
 5 {
 6     int t,n,a[100005];
 7     cin>>t;
 8     for(int k=1;k<=t;k++)
 9     {
10         cin>>n;
11         for(int i=1;i<=n;i++)
12             cin>>a[i];
13         int max,ans=-99999999,a1,a2;
14         for(int i=1;i<=n;i++)
15         {
16             max=0;
17             for(int j=i;j<=n;j++)
18             {
19                 max+=a[j];
20                 if(max>ans)
21                 {
22                    ans=max;
23                    a1=i;
24                    a2=j;
25                 }
26                 if(max<0) break;
27             }
28         }
29         printf("Case %d:\n",k);
30         printf("%d %d %d\n",ans,a1,a2);
31         if(k<t) cout<<endl;
32     }
33 }

 

 1 #include <cstdio>
 2 int main(){
 3     int t,cas=0;
 4     scanf("%d",&t);
 5     while(t--&&++cas)
 6     {
 7         if(cas!=1) printf("\n");
 8         printf("Case %d:\n",cas);
 9         int a[100001],f[100001]={0};
10         int max,maxl,maxr,l,r,n;
11         scanf("%d",&n);
12         l=1; max=-100000;
13         for(int i=1;i<=n;i++){
14             scanf("%d",&a[i]);
15             if(f[i-1]+a[i]<a[i]) { f[i]=a[i]; l=i; r=i; }
16             else{ f[i]=f[i-1]+a[i]; r=i; }
17             if(max<f[i]){ max=f[i]; maxl=l; maxr=r; }
18         } printf("%d %d %d\n",max,maxl,maxr);
19     } return 0;
20 }