【线性DP 子序列和】Max Sum（HDU-1003）

【线性DP 子序列和】Max Sum（HDU-1003）

题目

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

思路

求最大连续子序列和还要找出首尾区间
主要是累加sum
如果sum>max_sum那么更新区间并累加
如果sum<0那么直接舍弃sum设为0，开始下一个
主要是找首尾区间啊啊写了好几发全都蜜汁WA了

代码

个人过不了的两个代码

#include<bits/stdc++.h>
using namespace std;
int T,n;
const int N=1e5+10,INF=0x3f3f3f3f;
typedef long long ll;
int  a[N],f[N];
int main()
{
    scanf("%d",&T);
    for(int C=1;C<=T;C++)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);

        int st=1,ed=1,ans=-INF;
        int sum=0;
        int tmpl=st;
        for(int i=1;i<=N;i++)
        {
            sum+=a[i];
            if(sum>ans)
            {
                ans=sum;
                st=tmpl;
                ed=i;
            }
            if(sum<0)
            {
                sum=0;
                tmpl=i+1;
            }
        }
        printf("Case %d:\n%d %d %d\n",C,ans,st,ed);
        if(C<T)printf("\n");
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int T,n;
const int N=1e5+10,INF=0x3f3f3f3f;
typedef long long ll;
int  a[N],s[N],f[N];
int main()
{
    scanf("%d",&T);
    for(int C=1;C<=T;C++)
    {
        scanf("%d",&n);
        memset(s,0,sizeof (s));
        memset(s,0,sizeof (f));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            s[i]=s[i-1]+a[i];
        }
        int st=1,ed=1,ans=-INF;
        for(int j=1,i=1;j<=n&&i<=j;j++)
        {
            while(s[j]-s[i-1]<0&&i<j)i++;
            if(s[j]-s[i-1]>ans)
            {
            //    printf("gx %d~%d=~%d\n",i,j,s[j]-s[i-1]);
                ans=s[j]-s[i-1];
                st=i;ed=j;
            }
        }
        printf("Case %d:\n%d %d %d\n",C,ans,st,ed);
        if(C<T)printf("\n");
    }
    return 0;
}

看别的大佬的AC代码
分开找起点和终点

#include <stdio.h>
#include <string.h>
#include <math.h>
int main()
{
  int t,n,k=1;
  scanf("%d",&t);
  while(k<=t)
  {
    scanf("%d",&n);
    int a[100001],max=-1000,start=0,end=0,i,sum=0;
    for(i=1;i<=n;i++)
    {
      scanf("%d",&a[i]);
      sum+=a[i];
      if(sum>max)
      {
        max=sum;
        end=i;
      }
      if(sum<0)
        sum=0;
    }
    sum=0;
    for(i=end;i>0;i--)
    {
      sum+=a[i];
      if(sum==max)
        start=i;
    }
    printf("Case %d:\n",k);
    printf("%d %d %d\n",max,start,end);
    if(k!=t)
      printf("\n");
    k++;
  }
  return 0;
}