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
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思路
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状态转移方程是：sum[i]=sum[i-1]>0?:sum[i-1]+a[i]:a[i]
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源码
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#include<stdio.h>
#define min -99999999
int main ()
{
int i, n, m, sum, end, start, j, son, b, maxsum;
scanf("%d", &n) ;
for(i=1; i<=n; i++ )
{
scanf("%d", &m ) ;
start=0;
end=0 ;
maxsum=min ;
sum=0 ;
b=1 ;
for(j=1; j<=m; j++ )
{
scanf("%d", &son) ;
sum+=son ;
if(sum>maxsum)
{
maxsum=sum ;
start=b;
end=j ;
}
if(sum<0)
{
sum = 0;
b = j+1 ;
}
}
printf( "Case %d:\n", i) ;
printf( "%d %d %d\n", maxsum, start, end);
if(i!=n)
printf("\n") ;
}
}
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