hdu 1003 Max Sum 解题报告
链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003
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
简单的动态规划，最大子串和；
/*
******** hdu 1003 ***********
*/
/*
******** 琴心&剑胆 ***********
*/
/*
******** 2011/5/9 ***********
*/
#include
<
stdio.h
>
int
main(){
int
T;
scanf(
"
%d
"
,
&
T );
int
M
=
T;
while
( T
--
){
int
n;
scanf(
"
%d
"
,
&
n );
int
a[n],ts
=
0
,ti
=
0
,s
=
0
,bi
=
0
,bj
=
n
-
1
,f
=
0
,max
=-
32767
,mi;
for
(
int
i
=
0
;i
<
n;
++
i ){
scanf(
"
%d
"
,
&
a[i] );
if
(max
<
a[i]){
max
=
a[i];
mi
=
i;
}
if
( a[i]
<
0
)
f
++
;
ts
+=
a[i];
if
(ts
<
0
){
ts
=
0
;
ti
=
i
+
1
;
}
if
(s
<
ts){
s
=
ts;
bj
=
i;
bi
=
ti;
}
}
printf(
"
Case %d:\n
"
,M
-
T);
if
( f
==
n )
printf(
"
%d %d %d\n
"
,max,mi
+
1
,mi
+
1
);
else
printf(
"
%d %d %d\n
"
,s,bi
+
1
,bj
+
1
);
if
( T )
printf(
"
\n
"
);
}
}