Max Sum(最大连续子序列)

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

贴两个ac代码吧，心累（第一个摘抄网上）

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=500009;
const int INF=0x3f3f3f3f;
ll a[maxn];
int main()
{
    ll T,n,i,j,k;
    scanf("%lld",&T);
    for(k=1;k<=T;k++)
    {
        scanf("%lld",&n);
        for(i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        ll s=0,max1=-1e9,start=1,endd=1,t=1;
        for(i=1;i<=n;i++)
        {
            s+=a[i];
            if(max1<=s)
            {
                max1=s;
                endd=i;
                start=t;
            }
            if(s<0)
            {
                s=0;
                t=i+1;
 
            }
        }
        printf("Case %d:\n",k);
        printf("%lld %lld %lld\n",max1,start,endd);
        if(T!=k)
           puts("");
    }
    return 0;
}

都挺好理解的，跑一遍就能搞懂，主要还是理清楚思路吧

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<math.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
int sum[100010],a[100010];
int main()
{
    int t,k,x,y,flag;
    scanf("%d",&t);
    k=0;
    while(t--)
    {
        k++;
        memset(sum,0,sizeof(sum));
        int n,max1=-1001;
        int i,j;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=n;i++)
        {
            sum[i]=max(sum[i-1]+a[i],a[i]);
            if(sum[i]>max1)
            {
                max1=sum[i];
                x=i;
            }
        }
        flag=max1;
        for(i=x;i>=1;i--)
        {
            max1-=a[i];
            if(max1==0)
            {
                y=i;
                break;
            }
        }
        if(k!=1)
            printf("\nCase %d:\n%d %d %d\n",k,flag,y,x);
        if(k==1)
            printf("Case %d:\n%d %d %d\n",k,flag,y,x);
    }
    return 0;
}