杭电1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 183237    Accepted Submission(s): 42745

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

 

Author

Ignatius.L

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

注意考虑数据全为负数的情况！！！！

#include <stdio.h>int main(){    int t,n,i;    int a[100000];    int sum,max,start,end,s,e;    int id=0,flag=0,fu;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        fu=0;        for (i=0; i<n; i++)        {            scanf("%d",&a[i]);            if (a[i]<0)            {                fu++;            }        }        if (flag)        {            printf("\n");        }        flag=1;        printf("Case %d:\n",++id);        if (fu==n)//全为负        {            max=a[0];            s=0;            for(i=1; i<n; i++)            {                if (a[i]>max)                {                    max=a[i];                    s=i;                }            }            printf("%d %d %d\n",max,s+1, s+1);            continue;        }        sum=start=end=s=e=0;        max=-1000;        for (i=0; i<n; i++)        {            sum+=a[i];            if (sum>=0)            {                end=i;            }            else /**(sum<0)**/            {                sum=0;                start=i+1;                end=i+1;            }            if (sum>=max)            {                max=sum;                s=start;                e=end;            }        }        printf("%d %d %d\n",max,s+1,e+1);    }    return 0;}