题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 178388 Accepted Submission(s): 41628
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
Author
Ignatius.L
Recommend
题意 ： 给出一个序列，输出最大子序列和，简单的dp
dp数组储存的是以这个值结尾的最长的子序列和
dp[i] = max(dp[i-1]+num[i] , num[i]);
但是因为要保存起始和终止点的位置，所以可以用结构体来储存dp；
下面是代码
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 #define N 100005 6 #define ll long long 7 struct DP{ 8 ll sum; 9 int l; 10 int r; 11 bool operator < (const DP d) const 12 { 13 if(sum!=d.sum) return d.sum<sum; 14 else if(l!=d.l) return l<d.l; 15 else return r<d.r; 16 } 17 }dp[N]; 18 ll num[N]; 19 int main() 20 { 21 int T; 22 scanf("%d",&T); 23 for(int cnt = 0 ; cnt < T ; cnt++) 24 { 25 int n; 26 scanf("%d",&n); 27 for(int i = 0 ;i < n ;i++) 28 dp[i].sum = 0, dp[i].l = i,dp[i].r = i; 29 for(int i = 0 ;i < n ;i++) 30 { 31 scanf("%lld",&num[i]); 32 if(i==0) dp[i].sum = num[0],dp[i].l = 0,dp[i].r = 0; 33 34 else 35 { 36 if(dp[i-1].sum+num[i]>=num[i]) 37 { 38 dp[i].sum = dp[i-1].sum+num[i]; 39 dp[i].l = dp[i-1].l; 40 dp[i].r = i; 41 } 42 else 43 { 44 dp[i].sum = num[i]; 45 dp[i].l = i; 46 dp[i].r = i; 47 } 48 } 49 } 50 sort(dp,dp+n); 51 if(cnt!=0) puts(""); 52 printf("Case %d: ",cnt+1); 53 printf("%lld %d %d ",dp[0].sum,dp[0].l+1,dp[0].r+1); 54 } 55 return 0; 56 }