Max Sum
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
Analysis
状态转移方程 dp[i] = max( dp[i-1] + a[i] , a[i] )
1 #include <stdio.h>
2 #include <string.h>
3
4 int a[100002];
5
6 main()
7 {
8 int T,N,kkk=1,i,j;
9
10 scanf("%d",&T);
11 while(T--)
12 {
13 int max,max_s,max_e,sum,sum_s;
14 //memset
15 printf("Case %d:\n",kkk++);
16
17 scanf("%d",&N);
18 for(i=1;i<=N;i++)
19 {
20 scanf("%d",&a[i]);
21 }
22
23 max = sum = a[1];
24 max_s = max_e = sum_s = 1;
25
26 //printf("sum = %d max = %d\n",sum,max);
27 for(i=2;i<=N;i++)
28 {
29 if(sum + a[i] < a[i])
30 {
31 sum = a[i] ;
32 sum_s = i ;
33 }
34 else sum = sum + a[i] ;
35
36 if(sum > max)
37 {
38 max = sum ;
39 max_s = sum_s;
40 max_e = i;
41 }
42 //printf("sum = %d max = %d\n",sum,max);
43 }
44 printf("%d %d %d\n",max,max_s,max_e);
45 if(T)
46 printf("\n");
47 }
48 }