【HDU 1003】 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 257916    Accepted Submission(s): 61282

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

分析：典型的求最长字段和的题目，用时间复杂度为O(n)的算法写。

代码：

#include<cstdio>int a[100005],T,n;int main(){    scanf("%d",&T);    for (int tt=1;tt<=T;++tt)    {        int l=0,r=0,k=0,ans=-1000,now=1;        scanf("%d",&n);        for (int i=1;i<=n;++i) scanf("%d",&a[i]);        printf("Case %d:\n",tt);        for (int i=1;i<=n;++i)        {            k+=a[i];            if (k>ans) ans=k,l=now,r=i;            if (k<0) k=0,now=i+1;        }        printf("%d %d %d\n",ans,l,r);        if (tt!=T) puts("");    }    return 0;}