HDU1003——Max Sum（DP）

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

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

 

分析：

听说这是DP、我没研究过DP题目、这道题按照自己思路写的、中间有点小卡壳、看

了一些博客、借鉴了一下。听说这是DP的入门级别的题。如果各个子问题不是独立

的，不同的子问题的个数只是多项式量级，如果我们能够保存已经解决的子问题的答

案，而在需要的时候再找出已求得的答案，这样就可以避免大量的重复计算。由此而

来的基本思路是，用一个表记录所有已解决的子问题的答案，不管该问题以后是否被

用到，只要它被计算过，就将其结果填入表中——from LCY

——>就是总是把当前最大的数记录下来、然后扫描一遍。

#include<iostream>#include<string.h>#include<stdio.h>#include<ctype.h>#include<algorithm>#include<stack>#include<queue>#include<set>#include<math.h>#include<vector>#include<map>#include<deque>#include<list>using namespace std;int n,t;int i,j;int start,end;int a[100007];int w=1;int f(int p){    int max,ls,le;    int sum=max=-99999;    for(int i=0; i<p; i++)    {        if(sum<0)        {            if(a[i]>sum)            {                sum=a[i];                ls=le=i;                if(sum>max)                {                    max=sum;                    start=ls;                    end=le;                }            }        }        else        {            sum+=a[i];            le=i;            if(sum>max)            {                max=sum;                start=ls;                end=le;            }        }    }    return max;}int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0; i<n; i++)            scanf("%d",&a[i]);        int r=f(n);        printf("Case %d:\n",w++);        printf("%d %d %d\n",r,start+1,end+1);        if(t)            printf("\n");    }    return 0;}