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[ACM] hdu 1003 Max Sum（最大子段和模型）
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 135262 Accepted Submission(s): 31311
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
25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5
Sample Output
Case 1:14 1 4Case 2:7 1 6
Author
Ignatius.L
解题思路：
当所有数都为负数时，最大子段和为0.
int MaxSum(int num[],int n){ int sum=0,b=0; int i; for (i=1;i<=n;i++) { if(b>0) b+=num[i]; else b=num[i]; if(b>sum) sum=b; } return sum;}只求最大和，没有保存位置。
保存位置：start为起 end为末
int start=1,end=1,s=1,e=1; int sum=0,max=num[0];//不能让max=0 for(int i=1;i<=n;i++) { e=i; sum=sum+num[i]; if(max<sum) { max=sum; start=s; end=e; } if(sum<0) { s=i+1; sum=0; } }
本题需要保存起始位置。
代码：
#include <iostream>using namespace std;const int maxn=100002;int num[maxn];int n;int main(){ int t;cin>>t;int c=1; while(t--) { cin>>n; for(int i=1;i<=n;i++) cin>>num[i]; int start=1,end=1,s=1,e=1; int sum=0,max=num[0];//不能让max=0 for(int i=1;i<=n;i++) { e=i; sum=sum+num[i]; if(max<sum) { max=sum; start=s; end=e; } if(sum<0) { s=i+1; sum=0; } } cout<<"Case "<<c++<<":"<<endl; cout<<max<<" "<<start<<" "<<end<<endl; if(t) cout<<endl; } return 0;}
[ACM] hdu 1003 Max Sum（最大子段和模型）
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