A1007 Maximum Subsequence Sum (25) 动态规划

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原创最新推荐文章于 2026-05-22 12:39:43 发布 166 阅读 AI 写同款· GEO 优化 ›

PAT练习题(甲级) 1007 Maximum Subsequence Sum (25分)(简要理解动态规划)(Java实现) PAT练习题 1007 Maximum Subsequence Sum （25分）题干题干 Given a sequence of K integers { N1, N2, …, NK}. A continuous subsequence is defined to be { Ni, Ni+1, …, Nj} where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of 阅读详情

注意输出的是对应的数值，而不是下标

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=100010;
struct node
{
	int st;
	int ed;
    int data;
}a[maxn];
int d[maxn]={0};
bool cmp(node b,node c)
{
	if(b.data!=c.data)return b.data>c.data;
	else return b.ed<c.ed;
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&d[i]);
	}
	a[0].data=d[0];
	a[0].st=0;
	a[0].ed=0;
	for(int i=1;i<n;i++)
	{
		if(a[i-1].data>0)
		{
			a[i].data=a[i-1].data+d[i];
			a[i].st=a[i-1].st;
			a[i].ed=i;
		}
		else
		{
			a[i].data=d[i];
			a[i].st=i;
			a[i].ed=i;
		}
	}
	sort(a,a+n,cmp);
	if(a[0].data<0) printf("0 %d %d",d[0],d[n-1]);
	else printf("%d %d %d",a[0].data,d[a[0].st],d[a[0].ed]);
}