HDOJ-1003-Max Sum

HDOJ-1003

Max Sum*

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 342081 Accepted Submission(s): 81313

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

Author
Ignatius.L

ansi,ansj分别为从第几个数开始和结束。
做法就是dp（动态规划），dp[i+1]就是前i个数所能得到的最大和，如果前一个dp小于0，那么从这个数开始重新计算，用ii暂时存储新的ansi,如果其中某一个dp大于了maxx(最大的dp），那么就让maxx=这个dp，顺便让ansi与ansj的值刷新成ii和现在的j
ps:其实可以一边读一遍操作，懒得改了

#include 
using namespace std;
typedef long long ll;
int t,h,i,j,n,a[100005],ansi,ansj,maxx,ii;
int dp[100005];
void solve()
{
	scanf("%d",&n);
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);
	maxx=dp[0]=a[0];
	ii=ansj=ansi=1;
	for(i=0;i<n-1;i++)
	{
			if(dp[i]<0)
			{
				dp[i+1]=a[i+1];
				ii=i+2; 
			} 
			else
				dp[i+1]=dp[i]+a[i+1];
			if(dp[i+1]>maxx)
			{
				maxx=dp[i+1];
				ansj=i+2;
				ansi=ii;
			}
			//printf("%d ",dp[i+1]);
	}
	printf("Case %d:\n%d %d %d\n",h-t,maxx,ansi,ansj);
	if(t!=0)
		printf("\n");
}
int main()
{
	/*
	freopen("data.in","r",stdin);
	freopen("data.out","w",stdout);
	*/
	scanf("%d",&t);
	h=t;
	while(t--)
    	solve();
    return 0;
}