題目回顧（HDU-1003）：

Max Sum

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 源碼與解析 1 #include<iostream> 2 #include<algorithm> 3 using namespace std; 4 int main(){ 5 int T,N; 6 int a[100010]; 7 cin>>T; 8 for(int k=1;k<=T;k++){ 9 cin>>N; 10 for(int i=0;i<N;i++){ 11 cin>>a[i]; 12 } 13 int sum=0; 14 int ans=-10000; //如果設置的數字不夠小，出錯。例如sum=-500， 15 　 //初始ans=-10，sum<ans無法標記下一步位置。 16 int start,end; //標記起點和終點。 17 int temp=0; //temp要初始化起點位置。 18 for(int i=0;i<N;i++){ 19 if(sum>=0){ 20 sum+=a[i]; 21 }else{ 22 sum=a[i]; 23 temp=i; //暫時保存新起點位置。 24 } 25 26 if(sum>ans){ //如果發現更大的sum，則保存起點和終點。 27 ans=sum; 28 start=temp; 29 end=i; 30 } 31 } 32 cout<<"Case "<<k<<":"<<endl<<ans<<" "<<start+1<<" "<<end+1<<endl; 33 if(k<T){ //輸出格式的要求。 34 cout<<endl; 35 } 36 } 37 return 0; 38 }

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轉載于:https://www.cnblogs.com/orangecyh/p/9762270.html

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