01-复杂度2 Maximum Subsequence Sum (25分)

Given a sequence of $N1N_1N1, N2N_2N2, ..., NKN_KNK }. A continuous subsequence is defined to be { NiN_iNi, Ni+1N_{i+1}Ni+1, ..., NjN_jNj } where 1≤i≤j≤K1 le i le j le K1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.$

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

时间限制：200ms
内存限制：64MB
代码长度限制：16kB
判题程序：系统默认
作者：陈越
单位：浙江大学

#include<cstdio>
using namespace std;
int main()
{
    int k,csum=0,sum=-1,x=0,t=0,y=-1;
    int s[10001];
    scanf("%d",&k);
    for(int i=0;i<k;i++)scanf("%d",s+i);
    for(int i=0;i<k;i++){
        csum+=s[i];
        if(csum>sum){
            sum= ;x=t;y=i;
        }
        if(csum<0){
            csum=0;t=i+1;
        }
    }
    if(y==-1)   printf("0 %d %d
",s[0],s[k-1]);
    else printf("%d %d %d
",sum,s[x],s[y]);
}

t 表示成为下一个最大子序列的第一个节点

原文地址：https://www.cnblogs.com/acmtime/p/5905659.html

01-复杂度2 Maximum Subsequence Sum (25分)

Input Specification:

Output Specification:

Sample Input:

Sample Output:

题目判定

解题程序