1481:Maximum sum
- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
- Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
t1 t2 d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n } i=s1 j=s2
Your task is to calculate d(A).
- 输入
- The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case. - 输出
- Print exactly one line for each test case. The line should contain the integer d(A).
- 样例输入
-
1 10 1 -1 2 2 3 -3 4 -4 5 -5
- 样例输出
-
13 //最大连续和的变种,dp数组表示以i结尾的子串最大和,l数组表示i往左最大的一个和,后二者表示反过来
#include<iostream>#include<memory.h>using namespace std;int dp[50001], rdp[50001], l[50001], r[50001], data[50001];int main() { int T; scanf("%d", &T); while (T --) { int n; scanf("%d", &n); for (int i = 0; i < n; i ++) scanf("%d", &data[i]); dp[0] = data[0]; rdp[n - 1] = data[n - 1]; memset(l, 0, sizeof(l)); memset(r, 0, sizeof(r)); l[0] = dp[0]; r[n - 1] = rdp[n - 1]; for (int i = 1; i < n; i ++) { dp[i] = data[i]; if (dp[i - 1] > 0) dp[i] += dp[i - 1]; l[i] = max(l[i - 1], dp[i]); } for (int i = n - 2; i >= 0; i --) { rdp[i] = data[i]; if (rdp[i + 1] > 0) rdp[i] += rdp[i + 1]; r[i] = max(r[i + 1], rdp[i]); } int ans = -2147483646;//注意全为负数的情况 for (int i = 0; i < n - 1; i ++) ans = max(ans, l[i] + r[i + 1]); printf("%d\n", ans); }}
时间: 2024-08-11 05:46:23