Max Sum
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 11 Accepted Submission(s) : 5
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<stdio.h> #include<string.h> #define MAX 100000+10 #define INF 0x3f3f3f int main() { int n,m,j,i,k1,k2,t,sum; int x,y,sum1; int s[MAX]; scanf("%d",&t); m=1; while(t--) { memset(s,0,sizeof(s)); scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&s[i]); } x=y=k1=k2=0; sum=s[0]; sum1=sum; for(i=1;i<n;i++) { if(sum>=0) { y=i; sum+=s[i]; } else { x=i; y=i; sum=s[i]; } if(sum>sum1) { sum1=sum; k1=x; k2=y; } } printf("Case %d:\n",m++); printf("%d %d %d\n",sum1,k1+1,k2+1); if(t!=0) printf("\n"); } return 0; }