PAT 甲级 1007 Maximum Subsequence Sum

#include <iostream>
#include <cstdio>
#include <cstring>
const int maxn = 10050;
struct node
{
    int start,end;
    int sum;
    int value;
}dp[maxn];

int main() {
    //std::cout << "Hello, World!" << std::endl;
    int K;
    scanf("%d",&K);
    for (int i = 0; i < K; ++i) {
    scanf("%d",&dp[i].value);
    }
    int maxSum=dp[0].value;
    dp[0].sum=dp[0].value;
    dp[0].start=dp[0].end=0;
    for (int i = 1; i < K; ++i) {
        if(dp[i].value+dp[i-1].sum>dp[i].value){
            dp[i].sum=dp[i].value+dp[i-1].sum;
            dp[i].start=dp[i-1].start;
            dp[i].end=i;
        }else{
            dp[i].sum=dp[i].value;
            dp[i].start=dp[i].end=i;
        }
        maxSum=maxSum>dp[i].sum?maxSum:dp[i].sum;
    }
    if(maxSum<0){
        printf("0 %d %d
",dp[0].value,dp[K-1].value);
    }else{//注意输出下标最小的数值！！！
        printf("%d ",maxSum);
        for (int i = 0; i < K; ++i) {
            if(maxSum==dp[i].sum){
                printf("%d %d
",dp[dp[i].start].value,dp[dp[i].end].value);
                break;
            }
        }
    }
    return 0;
}

涉及动态规划，没学过打扰勿喷，个人理解就是下一节点保存上一节点的求和或者最下下标

ps.输出最小的首末数值