J - Max Sum
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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
//第一行代表有 t 组测试案例，问最大连续的和为多少，并输出第几位数到第几位数
//虽然，这是一道dp水题，但我从没做过，自己写出来感觉不错，62ms 比别人的慢了一倍，应该是我用了两次一重循环吧。
1 #include <stdio.h>
2
3 #define MAXN 100005
4
5 int num[MAXN];
6 int dp[MAXN];
7
8 int cmp(int a,int b)
9 {
10 return a>b?a:b;
11 }
12
13 int main()
14 {
15 int t;
16 int n,i,Case=0;
17 scanf("%d",&t);
18 while (t--)
19 {
20 scanf("%d",&n);
21 int max;
22 for (i=1;i<=n;i++)
23 {
24 if (i==1)
25 {
26 scanf("%d",&num[i]);
27 dp[i]=num[i];
28 max=i;
29 continue;
30 }
31 scanf("%d",&num[i]);
32 dp[i]=cmp(num[i],dp[i-1]+num[i]);
33 if (dp[i]>dp[max]) max=i;
34 }
35 int s=max;
36 for (i=max;i>=1;i--)
37 {
38 if (dp[i]>=0)
39 s=i;
40 if (dp[i]<0)
41 break;
42 }
43 printf("Case %d:
",++Case);
44 printf("%d %d %d
",dp[max],s,max);
45 if (t!=0) printf("
");
46 }
47 return 0;
48 }