Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49213 Accepted Submission(s): 10950
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
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1 #include<iostream> 2 #include<vector> 3 using namespace std; 4 5 #define P(x) \ 6 cout << #x " " << x << ":" << endl; 7 8 #define PE(max, beg, end) \ 9 cout << max << " " << beg+1 << " " << end+1 << endl; 10 11 int max_sum(int &beg, int &end, vector<int> ivec) 12 { 13 while(ivec[beg] < 0 && beg < ivec.size()) 14 { 15 //忽略首相的负数 16 beg++; 17 } 18 int max = 0, now = 0; 19 for(int i = beg; i < ivec.size(); ++i) 20 { 21 now += ivec[i]; 22 if(max < now) 23 { 24 max = now; 25 end = i; 26 } 27 } 28 return max; 29 } 30 int main() 31 { 32 int T, Case = 1; 33 cin >> T; 34 while(T--) 35 { 36 vector<int> ivec; 37 int n, x; 38 cin >> n; 39 while(n--) 40 { 41 cin >> x; 42 ivec.push_back(x); 43 } 44 int beg = 0, end = 0, now_beg = 0, now_end = 0, now = -1000, max = -1000; 45 for(; now_beg < ivec.size(); ++now_beg) 46 { 47 now = max_sum(now_beg, now_end, ivec); 48 if(max < now) 49 { 50 max = now; 51 beg = now_beg; 52 end = now_end; 53 } 54 } 55 P(Case); 56 Case++; 57 PE(max, beg, end); 58 if(T) 59 cout << endl; 60 } 61 return 0; 62 }
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1 #include <iostream> 2 using namespace std; 3 int main() 4 { 5 int T,N,num,startP,endP; 6 cin>>T; 7 for(int k=0;k<T;k++) 8 { 9 cin>>N; 10 int max=-1001,sum=0,temp=1; 11 for(int i=0;i<N;i++) 12 { 13 cin>>num; 14 sum+=num; 15 if(sum>max) 16 { 17 max=sum; 18 startP=temp; 19 endP=i+1; 20 } 21 if(sum<0) 22 { 23 sum=0; 24 temp=i+2; 25 } 26 } 27 cout<<"Case "<<k+1<<":"<<endl<<max<<" "<<startP<<" "<<endP<<endl; 28 if(k!=T-1) cout<<endl; 29 } 30 return 0; 31 }