最大子列和问题(同时输出有最大和的子列的首尾元素)【数据结构测试1.2】
2015-01-02 12:00
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题目:
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
PS:标明注意的部分部分比较容易忽略,一开始反复提交都是部分正确,就是因为少了这种特例的判断
测试程序:
以上程序测试数据:(a.txt)
6
-2 11 -4 13 -5 -2
10
-10 1 2 3 4 -5 -23 3 7 -21
6
5 -8 3 2 5 0
3
-1 -5 -2
3
-1 0 -2
1
10
提交到OJ的代码:
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
PS:标明注意的部分部分比较容易忽略,一开始反复提交都是部分正确,就是因为少了这种特例的判断
测试程序:
// maxSubSum.cpp : 定义控制台应用程序的入口点。 #include "stdafx.h" #include "iostream" #include "fstream" #include <vector> using namespace std; int str_pnt[2] = {0}, end_pnt = 0;//开始点,结束点 /* 修改最大子列和程序,记录有最大和的子列的首尾 */ long maxSubsum(vector<int> &a) { int maxSum = 0,thisSum = 0; for(int i = 0; i < a.size(); i++) { thisSum += a[i]; if(thisSum >= maxSum) { maxSum = thisSum; end_pnt = i;//每次更新max的时候更新结束点位置 str_pnt[0] = str_pnt[1]; } else if(thisSum == 0 && maxSum == 0)//此处注意:处理序列类似[-1 0 -2]的这种情况(此种情况应该输出0 0 0,无此条件则会输出0 -1 -1) { end_pnt = i;//每次更新max的时候更新结束点位置 str_pnt[0] = str_pnt[1]; } else if(thisSum < 0) { thisSum = 0; str_pnt[1] = i+1;//若子序列移动,更新开始点位置(保存此时位置) } } return maxSum; } int main() { int n = 0,i = 0; int flag = 0;//是否有非负数标志位 fstream cin("a.txt"); cin >> n; while(!cin.eof()) { vector<int> a(n); while(n--) { cin >> a[i]; if(a[i] >= 0) flag = 1; i++; } i = 0; cout << maxSubsum(a); if(! flag )//若整个序列均为负数 cout<<" "<<a[0]<<" "<<a[a.size()-1]<<endl; else cout <<" "<< a[str_pnt[0]] << " " << a[end_pnt] <<endl; cin >> n; //清零 str_pnt[0] = str_pnt[1] = 0; end_pnt = 0; a.clear(); } return 0; }
以上程序测试数据:(a.txt)
6
-2 11 -4 13 -5 -2
10
-10 1 2 3 4 -5 -23 3 7 -21
6
5 -8 3 2 5 0
3
-1 -5 -2
3
-1 0 -2
1
10
提交到OJ的代码:
int main() { int n = 0,i = 0; int flag = 0; cin >> n; vector<int> a(n); while(n--) { cin >> a[i]; if(a[i] >= 0) flag = 1; i++; } cout << maxSubsum(a); if(! flag )//若整个序列均为负数 cout<<" "<<a[0]<<" "<<a[a.size()-1]<<endl; else cout <<" "<< a[str_pnt[0]] << " " << a[end_pnt] <<endl; return 0; }
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