zoukankan      html  css  js  c++  java
  • PAT 1007 Maximum Subsequence Sum

    1007 Maximum Subsequence Sum (25 分)
     

    Given a sequence of K integers { N1​​, N2​​, ..., NK​​ }. A continuous subsequence is defined to be { Ni​​, Ni+1​​, ..., Nj​​ } where 1. 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 (≤). The second line contains Knumbers, 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

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    
    
    
    int main(){
        int n;
        cin >> n;
        int num[n];
        int flag = 1;
        for(int i=0;i < n;i++) {
            cin >> num[i];
            if(num[i] >= 0)flag = 0;
        }
        if(flag){
            cout << 0<<" " << num[0] <<" "<< num[n-1];
            return 0;
        }
    
    
        int sum = 0;
        int maxsum = 0;
        int j=0,k=0;
        int movej=0;
    
        for(int i=0;i < n;i++){
            sum += num[i];
            if(sum < 0) {sum = 0;movej=i+1;}
            if(maxsum < sum){
                j = movej;k = i;
                maxsum = sum;
            }
        }
        if(num[j] < 0)num[j] = 0;
        if(num[k] < 0)num[k] = 0;
    
        cout << maxsum << " " << num[j] << " " << num[k];
    
    
        return 0;
    }

    看错输出,应该是输出数字的,我输出下标了,还过了2个数据点??

    然后边界条件有个

    3
    -1 -1 0

    应该是输出0 0 0

  • 相关阅读:
    showModalDialog 页面上GridView的分页问题
    Ibatisnet Quick Start
    frame,iframe,frameset之间的关系与区别
    JS控制彈出窗口
    注册Dll
    ibatis动态查询条件
    前端性能优化之文件按需异步加载
    Redis 实践笔记
    性能测试:Redis千万级的数据量的性能测试
    js控制文本框只能输入中文、英文、数字与指定特殊符号.
  • 原文地址:https://www.cnblogs.com/cunyusup/p/10680730.html
Copyright © 2011-2022 走看看