HDU1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 246052    Accepted Submission(s): 58108

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
//题目是要求计算给定数列按顺序相加最后的结果的最大值，并输出结果以及最大子数列的起始和终止位置
//直接寻找起始元素>=0 以及终止元素>=0 最后的结果就是最大的
#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;int main(int argc, const char * argv[]) {    int m,n,jishu=1;    scanf("%d",&n);    int a[100001];    while(n--)    {        scanf("%d" ,&m);        int count = 1,sum = 0;        int first = 1,end = 1,max=-1001;        for(int i = 1;i <= m; ++i)        {            scanf("%d",&a[i]);            sum += a[i];            if(sum > max)            {                max = sum;                first = count;                end = i;            }            if(sum < 0)            {                count = i+1;                sum = 0;            }        }        cout << "Case " << jishu++ << ":"<< endl;        cout << max << " " << first << " " << end << endl;        if(n > 0)            cout << endl;    }    return 0;}