Max Sum

Vjudge

• 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 				
  

• Output

  Case 1:
  14 1 4
  
  Case 2:
  7 1 6
  

  原思路：直接DP，用一维数组，否则会爆，但还是会超时
  超时代码如下

  点击查看代码

  #include <iostream>
  #include <cstdio>
  #include <algorithm>
  #include <math.h>
  #include <string.h>
  using namespace std;
  int t,n,a[100005],dp[100005],k;
  void solve()
  {
  	cin>>n;
  	memset(a,0,sizeof(a));
  	memset(dp,0,sizeof(dp));
  //	memset(vis,0,sizeof(vis));
  	for(int i=1;i<=n;i++)
  	{
  		cin>>a[i];
  	}
  	int ans=0,st=0,en=0;
  	for(int i=1;i<=n;i++)
  	{
  		for(int j=i;j<=n;j++)
  		{
  			if(dp[j]<dp[j-1]+a[j])
  			{
  				dp[j]=dp[j-1]+a[j];
  			}
  			//dp[j]=max(dp[j],dp[j-1]+a[j]);
  			if(ans<dp[j])
  			{
  				ans=dp[j];
  				st=i,en=j;
  			}
  
  //			cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
  		}
  	}
  	cout<<"Case "<<k<<":"<<endl;
  	cout<<ans<<" "<<st<<" "<<en<<endl;
  	if(t)cout<<endl;
  }
  int main()
  {
  	ios_base::sync_with_stdio(false);
  	cin.tie(0);cout.tie(0);
  	cin>>t;
  	while(t--)
  	{
  		k++;
  		solve();
  	}
  	return 0;
  

}
```

O(n^2)的复杂度显然会超时，所以我们可以用前缀和维护，代替数组