题目描述:

Maxsum

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

思想

用summax变量存储从开始到当前的最大连续和,用k这个变量表示求累加和的开始位置,从第一个数开始遍历,求累加和,当此时累加和大于summax时,就要更新s_id和end_id,并更新summax,如果当前累加和<0了,则需要重新开始算,把当前位置赋值给k,并更新sum为0

代码:

#include<stdio.h>
#include<iostream>
using namespace std;
int arr[100010];
int main(){
  int t,cnt=0;
  cin>>t;
  while(t--){
    int n,s_id=0,e_id=0,sum=0,summax=-10000,k=1;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
      scanf("%d",&arr[i]);
    }
    for(int i=1;i<=n;i++){
      sum+=arr[i];
      if(sum>summax){
        summax=sum;
        s_id=k;
        e_id=i;
      }
      if(sum<0){
        sum=0;
        k=i+1;
      }
    }
    printf("Case %d:\n%d %d %d\n",++cnt,summax,s_id,e_id);
    if(t!=0) printf("\n");
  }	
}

第二种思想

时间复杂度为O(n)的方法:

对于每一个数,将这个数与前几个数的和和下一个数进行比较,如果前者大的话,就取前面的那个,否则就从下一个数开始。这样保证每次加的数的和都是最大的(如果不是最大的,就从下一个数开始重新计算了),所以,特别神奇吧。

代码:

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
const int MAXN=50050;
int n,ans=-2147483647,dp[MAXN],a[MAXN]; 
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++) 
    {
        scanf("%d",&a[i]);
        dp[i]=max(dp[i-1]+a[i],a[i]); //关键
        ans=max(ans,dp[i]);
    }
    printf("%d\n",ans);
    return 0;
} 

一个好奇的人