HDU-1003: Max Sum（最大连续子序列和）

DP

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

题意： t组数据，每组输入n再输入n个数，求最大连续子序列和，以及子序列的开头和结尾。

分析：若前面数的和大于等于0，那么现在这个数加上前面数的和肯定比自己本身更大或者等于本身，因此连续的子序列就要加上现在这个数，否则就不加。
sum[i]是以第i个数结尾的最大子序列的和，a[i]是原数组。
状态转移方程为：sum[i] = max{sum[i-1] + a[i], a[i]};
并且这道题需要找出序列的开头和结尾的位置，结尾的位置我们知道，因为已知以第i个数结尾最大的子序列和。问题是要找到开头。我们用数组st[i]存开头，如果前面序列的和大于0了，那么现在这个数还是纳入前面那个序列当中，所以st[i]=st[i-1]，否则它这个序列就以自己开头，因此st[i]=i。

• 完整代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int t,n,i,a[111111],max,st[111111],e,tt=0;
    scanf("%d",&t);
    int sb=t;
    while(t--)
    {
        tt++;
        memset(st,0,sizeof(st));
        memset(a,0,sizeof(a));
        max=-99999999;
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        st[0]=0;
        for(i=1;i<n;i++)
        {
            if(a[i-1]>=0)
            {
                a[i]+=a[i-1];
                st[i]=st[i-1];
            }
            else
                st[i]=i;
        }
        for(i=0;i<n;i++)
            if(a[i]>max)
            {
                max=a[i];
                e=i;
            }
        printf("Case %d:\n%d %d %d\n",tt,max,st[e]+1,e+1);
        if(tt!=sb)
            printf("\n");
    }
    return 0;
}