HDU 1003(动态规划-最大连续区间和)

ゝ一世哀愁。 2022-06-06 00:58 208阅读 0赞

问题描述:

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

题目题意:给一个数组，求最大连续区间和！

代码如下:

#include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    
    const int maxn=1e5+1000;
    int k,sum,ans,s,e,anss,anse,a[maxn];
    
    int main()
    {
        int t,icase=0;
        scanf("%d",&t);
        while (t--) {
            scanf("%d",&k);
            for (int i=1;i<=k;i++)
                scanf("%d",&a[i]);
            sum=ans=a[1];
            s=e=anss=anse=1;
            for (int i=2;i<=k;i++) {
                if (sum>=0) {
                    sum+=a[i];
                    e=i;
                }
                if (sum<0) {//如果前缀和小于0,那么另起起点肯定更优
                    sum=a[i];
                    s=e=i;
                }
                if (sum>ans) {
                    ans=sum;
                    anss=s;
                    anse=e;
                }
            }
            printf("Case %d:\n",++icase);
            printf("%d %d %d\n",ans,anss,anse);
            if (t) printf("\n");
        }
        return 0;
    }