Hdu 1003 Max Sum
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Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 250730 Accepted Submission(s): 59371
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#include<stdio.h>int main(){ int i,ca=1,t,s,e,n,x,now,before,max; scanf("%d",&t); while(t--){ scanf("%d",&n); for(i=1;i<=n;i++){ scanf("%d",&now); if(i==1){ max=before=now; x=s=e=1; } else { if(now>now+before){ before=now; x=i; } else before+=now; } if(before>max) max=before,s=x,e=i; } printf("Case %d:\n%d %d %d\n",ca++,max,s,e); if(t)printf("\n"); } return 0;}
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