HDU 1003

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串；

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]);

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    int t,n,a[100005];
    cin>>t;
    for(int k=1;k<=t;k++)
    {
        cin>>n;
        for(int i=1;i<=n;i++)
            cin>>a[i];
        int max,ans=-99999999,a1,a2;
        for(int i=1;i<=n;i++)
        {
            max=0;
            for(int j=i;j<=n;j++)
            {
                max+=a[j];
                if(max>ans)
                {
                   ans=max;
                   a1=i;
                   a2=j;
                }
                if(max<0) break;//显然跳过处理
            }
        }
        printf("Case %d:
",k);
        printf("%d %d %d
",ans,a1,a2);
        if(k<t) cout<<endl;
    }
}

#include <cstdio>
int main(){
    int t,cas=0;
    scanf("%d",&t);
    while(t--&&++cas)
    {
        if(cas!=1) printf("
");
        printf("Case %d:
",cas);
        int a[100001],f[100001]={0};
        int max,maxl,maxr,l,r,n;
        scanf("%d",&n);
        l=1; max=-100000;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            if(f[i-1]+a[i]<a[i]) { f[i]=a[i]; l=i; r=i; }
            else{ f[i]=f[i-1]+a[i]; r=i; }
            if(max<f[i]){ max=f[i]; maxl=l; maxr=r; }
        } printf("%d %d %d
",max,maxl,maxr);
    } return 0;
}