Max Sum -- hdu -- 1003

標籤：

連結：

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

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 180817    Accepted Submission(s): 42261

Problem DescriptionGiven 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. 

 

InputThe 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). 

 

OutputFor 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 Input25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5 

 

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

 

 

5 6 -1 5 4 -7
6 5 10 14 7  
Case 1:
14 1 4

7 0 6 -1 1 -6 7 -5
0 6 5 6 0 7 2
Case 2:
7 1 6

 

代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <queue>using namespace std;#define N 110000#define INF 0xffffffint sum[N];int main(){    int t, iCase=1;    scanf("%d", &t);    while(t--)    {        int i, n, Max=-INF, Min=INF, a, L, Left, Right;        scanf("%d", &n);        for(i=1; i<=n; i++)        {            scanf("%d", &a);            sum[i] = sum[i-1]+a;        }

/// Left = Right = 1;        for(i=1; i<=n; i++)        {
           ///找出最小的sum[i]           if(sum[i-1]<Min)           {               Min = sum[i-1];               L = i;           }           ///找出最大的Max           if(sum[i]-Min>Max)           {               Max = sum[i]-Min;               Left = L;               Right = i;           }        }        printf("Case %d:\n", iCase++);        printf("%d %d %d\n", Max, Left, Right);        if(t) printf("\n");    }    return 0;}

 

Max Sum -- hdu -- 1003