（HDU - 1003 ）Max Sum

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.

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

Case 1:
14 1 4

Case 2:
7 1 6

这就是模板题目，可能。需要注意就是在头节点的更换上面，不能在每一次更新最大字段和的时候就更新头head，很重要，需要更换头节点的时候必须满足ans>sum(ans为当前位置向前的最大字段，sum为这个位置之前的最大子段和) 的时候。这就需要用一个temp来保存当sum需要更换的时候的头节点变化。

AC代码：

#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e5+10;
int dp[N],aa[N];

int main(){
    int t,num=0,n;
    cin>>t;
    while(t--){
        cin>>n;
        for(int i=1;i<=n;i++) cin>>aa[i];
        int ans=aa[1],temp=1,head=1,tail=1;
        for(int i=1;i<=n;i++){
            if(dp[i-1]+aa[i]<aa[i]) temp=i;
            dp[i]=max(dp[i-1]+aa[i],aa[i]);
            if(dp[i]>ans){
                head=temp;
                tail=i;
            }
            ans=max(ans,dp[i]);
        }
        printf("Case %d:\n%d %d %d\n",++num,ans,head,tail);
        if(t)cout<<endl;
    }
    return 0;
}

标签：HDU,sequence,int,Max,sum,head,ans,line,1003
来源： https://www.cnblogs.com/jerrytangcaicai/p/16271074.html