专题三1001

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本文解析了一个经典的编程问题——如何寻找给定序列中拥有最大和的子序列及其位置。通过一个直观且易于理解的算法实现，展示了如何一步步找出最优解。

*题意：求给定序列的子序列的最大和，并输出最大和子序列的起始位置和结束位置

*解题思路：首先，这题是一道水题，首先将第一位数的和记为最大和以及sum，也记作起始位置和终止位置，如果sum加上后面一位数小于后面一个数本身，则sum=这个数，起始位置与终止位置也是这个数，反之，则sum+=这个数，终止位置变成这个数，然后比较sum与max，如果sum>max，则max=sum,就这样依次向后，最终就能求出最大和以及初始位置和结束为止

*感想：第一次提交，因为输出格式上 case后面漏了个：,结果提示我答案错误，我还以为是逻辑写错了，找了半天也没找到，又运行了一边程序，结果才发现输出有错误，也是醉了，以后写完的程序一定要在本地运行一遍再交。

*源码

# include <iostream>
using namespace std;
int main()
{
int T,n,i,j,ans,start,end,p,sum,a[100001];
cin>>T;
for(i=1;i<=T;i++)
{
if(i!=1)
cout<<endl;
cin>>n;
for(j=1;j<=n;j++)
cin>>a[j];
ans=0;
start=end=p=1;
ans=sum=a[1];
for(j=2;j<=n;j++)
{
if(sum<0)
{
sum=a[j];
p=j;
}
else
{
sum+=a[j];
}
if(sum>ans)
{
ans=sum;
start=p;
end=j;
}
}
cout<<"Case "<<i<<":"<<endl;
cout<<ans<<" "<<start<<" "<<end<<endl;
}
return 0;
}

Problem A

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 148 Accepted Submission(s) : 23

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.<br>

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).<br>

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.<br>

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

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