HDU__1003Max Sum（最大连续子序列和）

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

算是比较明显的DP题,很容易推出状态转移方程：sum[i] = max{sum[i-1]+a[i],a[i]}.

但是DP的代码也有好坏之分，我一开始就想麻烦了，觉得还得存起始下标和终点下标是不是得用个结构体啊，结果打完才发现，用不用结构体，dp的时候都得优化sum,左右下标，所以嘿嘿，第二遍才感觉正式的做出来了

比较好的（二改后）

#include<iostream>#include<vector>#include<algorithm>using namespace std;int putcnt = 0;int main(){int t,n,a,b,sum,retmax,l,r,L,R;cin>>t;while(t--){cin>>n;for(int i = 1;i <= n;i++){cin>>a; if(i == 1){sum = a;retmax = a;l = L = i;r = R = i;}else{if(b + a >= a){sum += a;r = i;} else{sum = a;l = i;r = i;}if(sum > retmax){retmax = sum;L = l;R = r;}}b = sum;//cout<<b<<endl;}printf("Case %d:\n",++putcnt);cout<<retmax<<" "<<L<<" "<<R<<endl;if(t >= 1){cout<<"\n";}}return 0; } 

一开始（麻烦的代码）

#include<iostream>#include<vector>#include<algorithm>using namespace std;struct node{int l;int r;int retmax;}findmax[100005];int putcnt = 0;int main(){int t,n,a,b,ret,retflag,l,r;cin>>t;while(t--){cin>>n;for(int i = 1;i <= n;i++){cin>>a; if(i == 1){findmax[i].l = 1;findmax[i].r = 1;findmax[i].retmax = a;ret = i;retflag = a;}else{if(b + a >= a){findmax[i].retmax = b + a;findmax[i].l = findmax[i - 1].l;findmax[i].r = i;} else{findmax[i].retmax = a;findmax[i].l = i;findmax[i].r = i;}if(findmax[i].retmax > retflag){retflag = findmax[i].retmax;ret = i;}}b = findmax[i].retmax;//cout<<b<<endl;}printf("Case %d:\n",++putcnt);cout<<findmax[ret].retmax<<" "<<findmax[ret].l<<" "<<findmax[ret].r<<endl;if(t >= 1){cout<<"\n";}}return 0; }