1 题目

1007 Maximum Subsequence Sum (25分)
Given a sequence of K integers { N​1, N2, …, NK​​ }. A continuous subsequence is defined to be { Ni​​ , N​i+1​​ , …, N​j​​ } 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 -21Sample Output:
10 1 4

2 解析

2.1 题意

求最大连续子序列的序列和以及序列的起始元素和结束元素

2.2 思路

用动态规划求解最大连续子序列的序列和

• 1 用dp[i]表示A[i]结尾的连续序列最大和（A[i]必须为连续序列的末尾）【不然就会产生多个相同的dp[i]】
  以-2 11 -4 13 -5 -2为例子（下标从0开始）
  dp[3] = 20(11+(-4) + 13)
  dp[4] = 15(11+(-4) + 13+(-5)), 如果A[i]不为连续序列的末尾，则值就会为20
• 2 dp为必须以A[i]为连续序列的结尾的序列，最大和有两种情况
  最大序列和只有一个元素，即以A[i]开始，以A[i]结尾
  最大序列和有多个元素，从前面某出A[p]开始(p<i)，一直到A[i]结尾。
• 3 因此状态转移方程为：
  dp[i] = max (A[i], d[i -1] + A[i]);
• 4 这个方程只与i和i前面的元素有关，且边界dp[0] = A[0],由此从小到大枚举i，来得到全部的数组dp

序列和的结束元素就是dp[i]最大的i；

序列和的起始元素求解为

• 当A[i] > d[i -1] + A[i],起始元素数组satrt[i] = i
• 否则(A[i] < d[i -1] + A[i])，start[i] = start[i - 1]

3 参考代码

#include <cstdio>
#include <algorithm>using std::max;const int MAXN = 10010;
int dp[MAXN];
int A[MAXN];
int start[MAXN];//记录dp[i]的连续序列从a的哪一个元素开始int main(int argc, char const *argv[])
{int n;scanf("%d", &n);int flag = false;for (int i = 0; i < n; ++i){scanf("%d", &A[i]);if(A[i] >= 0){flag = true;}}if(flag == false){//如果全部是负数printf("%d %d %d\n", 0, A[0], A[n - 1]);}else{dp[0] = A[0];start[0] = 0;for (int i = 1; i < n; ++i){if(A[i] > dp[i -1] + A[i]){dp[i] = A[i];start[i] = i;}else {dp[i] = dp[i -1] + A[i];start[i] = start[i-1];}}int index = 0;for (int i = 1; i < n; ++i){if(dp[index] < dp[i]){index = i;}}printf("%d %d %d\n", dp[index], A[start[index]], A[index]);}return 0;
}