HDU1003MAX SUM (动态规划求最大子序列的和)

程序员勇刚技术 2022年11月21日

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 206582 Accepted Submission(s):
48294

Problem DescriptionGiven 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. InputThe 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). OutputFor 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 Input2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5 Sample OutputCase 1:
14 1 4

Case 2:
7 1 6

AuthorIgnatius.L RecommendWe have carefully selected several similar problems for
you: 1176 1087 1069 1058 1159 题意：求连续子序列和的最大值。题解：注意这道题要求是输出第一个符合条件的序列，而且存在负数，样例之间要用空行分隔。

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+;
int a[maxn],n;
int main()
{
    int t,cas=;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int sum=,ans=-;
        int s=,e=,k=;
        for(int i=;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
            if(sum>ans)
            {
                s=k;
                e=i;
                ans=sum;
            }
            if(sum<) //0的意义就是这段数做的是负功
            {
                sum=;
                k=i+;
            }
        }
        printf("Case %d:\n",cas++);
        printf("%d %d %d\n",ans,s,e);
        if(t>) puts("");
    }
    return ;
}

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+;
int a[N],n;
int main()
{
    int t,cas=;
    cin>>t;
    while(t--)
    {
        cin>>n;
        int sum=,mx=-,s=,e=,ts=;
        for(int i=;i<=n;i++)
        {
            cin>>a[i];
            sum+=a[i];
            if(sum>mx)
            {
                mx=sum;
                s=ts;
                e=i;
            }
            if(sum<)
            {
                sum=;
                ts=i+;
            }
        }
        printf("Case %d:\n",cas++);
        printf("%d %d %d\n",mx,s,e);
        if(t) printf("\n");
    }
    return ;
}
/*
100
2 1 2
1 1
3 -1 1 2
2 -7 3
*/