zoukankan      html  css  js  c++  java
  • A1007. Maximum Subsequence Sum (25)

    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 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 K (<= 10000). The second line contains K numbers, separated by a space.

    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 i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

    Sample Input:

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

    Sample Output:

    10 1 4
     1 // hahaha.cpp : 定义控制台应用程序的入口点。
     2 //
     3 
     4 #include <stdafx.h>
     5 #include <stdio.h>
     6 #include <iostream>
     7 #include <vector>
     8 #include <map>
     9 #include <string>
    10 #include <cstdio>
    11 #include <set>
    12 #include <algorithm>
    13 #include <string.h>
    14 using namespace std;
    15 
    16 const int maxn=10010;
    17 int n;
    18 int a[maxn];
    19 
    20 
    21 int main()
    22 {
    23    scanf("%d",&n);
    24    bool flag=false;
    25    for(int i=0;i<n;i++)
    26        {
    27        scanf("%d",&a[i]);
    28        if(a[i]>=0)flag=true;
    29        }
    30    if(flag==false)
    31        {
    32        printf("0 %d %d",a[0],a[n-1]);
    33        return 0;
    34        }
    35    int dp[maxn];
    36    int s[maxn]={0};
    37    dp[0]=a[0];
    38    for(int i=1;i<n;i++)
    39        {
    40         if(dp[i-1]+a[i]>a[i])
    41             {
    42             dp[i]=dp[i-1]+a[i];
    43             s[i]=s[i-1];
    44             }else
    45             {
    46             dp[i]=a[i];
    47             s[i]=i;
    48             }
    49        }
    50    int k=0;
    51    for(int i=1;i<n;i++)
    52        {
    53        if(dp[i]>dp[k])
    54            {
    55            k=i;
    56            } 
    57        }
    58    printf("%d %d %d",dp[k],a[s[k]],a[k]);
    59     return 0;
    60 }
  • 相关阅读:
    第一阶段冲刺总结
    读书笔记之梦断代码1
    第二次站立会议5
    第二次站立会议4
    第二次站立会议3
    第二次站立会议2
    Java学习
    项目冲刺第十天
    项目冲刺第九天
    项目冲刺第八天
  • 原文地址:https://www.cnblogs.com/ligen/p/4342409.html
Copyright © 2011-2022 走看看