Max Sum

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

Total Submission(s): 135262    Accepted Submission(s): 31311

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

 

 

Author

 

Ignatius.L

 

解题思路：

当全部数都为负数时，最大子段和为0.

int MaxSum(int num[],int n){    int sum=0,b=0;    int i;    for (i=1;i<=n;i++)    {        if(b>0)            b+=num[i];        else            b=num[i];        if(b>sum)            sum=b;    }    return sum;}

仅仅求最大和，没有保存位置。

保存位置：start为起 end为末

int start=1,end=1,s=1,e=1;       int sum=0,max=num[1];//不能让max=0        for(int i=1;i<=n;i++)        {            e=i;            sum=sum+num[i];            if(max

 

本题须要保存起始位置。以下代码假设全是负数，输出最小的那个位置

代码：

#include 
     
      using namespace std;const int maxn=100002;int num[maxn];int n;int main(){    int t;cin>>t;int c=1;    while(t--)    {        cin>>n;        for(int i=1;i<=n;i++)            cin>>num[i];        int start=1,end=1,s=1,e=1;//这里start end一定要赋值为1        int sum=0,max=num[1];//不能让max=0        for(int i=1;i<=n;i++)        {            e=i;            sum=sum+num[i];            if(max