HDU1003 Max Sum【基礎DP 最大子段和】
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 302243 Accepted Submission(s): 71722
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
問題連結:HDU1003 Max Sum
問題描述:給你一段整數序列(可能包含負數),求子段和最大的子段,輸出最大和及其對應區間,如果有多個輸出第一個。
解題思路:最大子段和,基礎DP
AC的C++程式:
#include<iostream> #include<algorithm> using namespace std; const int N=100010; int a[N]; int main() { int T,n; scanf("%d",&T); for(int t=1;t<=T;t++){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); int l=1,l2=1,r=1,ans=-1001,sum=0; for(int i=1;i<=n;i++){ sum+=a[i]; if(sum>ans){ ans=sum; l=l2; r=i; } if(sum<0){ sum=0; l2=i+1; } } printf("Case %d:\n%d %d %d\n",t,ans,l,r); if(t!=T) printf("\n"); } return 0; }