hdu1003
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 112856 Accepted Submission(s): 26084
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
#include<iostream> #include<cstring> #include<string> using namespace std; int main() { int n,h = 1; scanf("%d",&n); while(n--) { int m,i,j,a[100010],start,k,end,max = -10010,sum=0; scanf("%d",&m); for(i=1;i<=m;i++) { scanf("%d",&a[i]); } for(i=1,k=1;i<=m;i++) { sum += a[i]; if(sum>max) { max = sum; start = k; end = i; } if(sum<0) { sum = 0; k = i+1; } } printf("Case %d:\n",h++); printf("%d %d %d\n",max,start,end); printf("\n"); } return 0; }