HDU 1003 Max Sum

最新推荐文章于 2022-03-03 08:27:19 发布

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最新推荐文章于 2022-03-03 08:27:19 发布

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本文详细介绍了一种求解子序列最大和的经典算法，通过动态规划的方法，有效地解决了给定整数序列中寻找具有最大和的连续子序列的问题。文章通过具体示例解释了算法的实现过程，并提供了一个完整的C++代码示例，帮助读者理解如何从输入的整数序列中计算最大和、子序列的起始位置和结束位置。

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int a[100002],b[100002];
    int T,j=0,N;
    cin>>T;
    while(T--)
    {
        int sum=0,max0=-1001,eend=0,begain=0;
        cin>>N;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=1; i<=N; i++)
        {
            cin>>a[i];
            b[i]=max(b[i-1]+a[i],a[i]);
            if(max0<b[i])
            {
                max0=b[i];
                eend=i;
            }
        }
        for(int k=eend; k>=1; k--)
        {
            sum+=a[k];
            if(sum==max0)
            {
                begain=k;
            }
        }
        printf("Case %d:\n",++j);
        printf("%d %d %d\n",max0,begain,eend);
        if(T)
            cout<<endl;
    }
    return 0;
}