Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 102625 Accepted Submission(s): 23592
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
代碼：
#include<stdio.h>
#define INT 0x7fffffff
int main()
{
int t,n,i,a,j;
int sum,MAX_SUM ;
int star,end,pos1;
scanf("%d",&t);
for(j=1;j<=t;j++)
{
sum=0;
star=1;
pos1=1;
MAX_SUM=-INT; //將最大值的初值設置爲最小。
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a);
if(sum<0)
{
sum=a; //如果該項與前幾項和sum的和的值小於0，則應從該項的下一項重新開始計數。
pos1=i;
}
else
sum+=a;
if(sum>MAX_SUM)
{
MAX_SUM=sum;
star=pos1;
end=i;
}
}
if(j!=t)
printf("Case %d:\n%d %d %d\n\n",j,MAX_SUM,star,end);
else
printf("Case %d:\n%d %d %d\n", j,MAX_SUM,star,end);
}
return 0;
}