标签：ace 序列 contains iostream acm between std pid 最大值
HDU1003
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
求最大子序列的和，及其左右开始下标。
甚是惭愧，现在是10月份，此题我在3月份已经在HDU上交过并AC了，但是没写题解，导致现在题出来又不会，现在必须补上。
此前我想的是，遇到a[i]<0时，重新更新sum值，但是后来一想不对，比如这个 : 2 -1 100 更新sum的话,最大只能加到100而不是101.
初始用dp保存输入的每个数字，而后的过程中，dp[i]为以 i 位置为结尾的最大值。如果走的过程中,dp[i-1]<0了，即累加出现<0的情况，那么就没必要再加了，因为再加就会减小。所以要从dp[i]重新累加，记录到此时的左端点 k=i; 每次for更新一次最大值(dp[i]>max)，l=k,r=i;
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1e5+10;
int dp[maxn]; //maxx=max{dp[i-1]+a[i],a[i]}
int main()
{
int t;
cin>>t;
int mid=t;
int ac=1;
while(t--)
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
cin>>dp[i];
int maxx=dp[1];
int l=1,r=1;
int k=1;
for(int i=2;i<=n;i++)
{
// cout<<dp[i-1]<<endl;
if(dp[i-1]>=0)
{
dp[i]+=dp[i-1];
}
else
{
k=i;
}
if(dp[i]>maxx)
{
maxx=dp[i];
l=k;
r=i;
}
}
printf("Case %d:\n",ac);
printf("%d %d %d\n",maxx,l,r);
if(ac!=mid)
cout<<endl;
ac++;
}
return 0;
}
注意输出，最后一句。每个样例之间输出一个空行，但是注意结尾不能有空行！
最大连续子序列和，以及开始、结束下标(HDU 1003)
标签：ace 序列 contains iostream acm between std pid 最大值
原文地址：https://www.cnblogs.com/liyexin/p/11693281.html
