#include <iostream>
#include <algorithm>
using namespace std;
struct node {
int h, j, m, w;
node(const int _h, const int _j, const int _m, const int _w) : h(_h), j(_j), m(_m), w(_w) {}
node operator+(const node &o) const {
return node(max(h, w + o.h), max(m + o.w, o.m), w + o.w, max(max(j, o.j), m + o.h));
}
};
node solve1(int h, int m) {
int j = (h + m) >> 1;
return solve1(h, j) + solve1(j + 1, m);
}
int solve2(int h, int m) {
int j = (h + m) >> 1;
if (h > m) return -1;
if (h == m) return max(a[h], 0);
int wht = 0, wmt = 0;
int wh = 0, wm = 0;
for (int i = j; i >= h; i--) {
wht += a[i];
wh = max(wh, wht);
}
for (int i = j + 1; i <= m; i++) {
wmt += a[i];
wm = max(wm, wmt);
}
return max(max(solve2(h, j), solve2(j + 1, m)), wh + wm);
}
int main() {
int n;
cin >> n;
int a[1010];
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
cout << solve2(1, n) << endl;
cout << solve1(1, n).j << endl;
return 0;
}
solve (1,n) 的时间复杂度为()。
A.Θ(logn) B.Θ(n) C.θ(nlogn) D.θ(n^2)。
下面这题不是球最大子串和吗?
输入 “10 -32 100 -89 -4 -594”,输出的第二行为()。
答案17,我24
为了防止再被*,我已迁址学术走一趟,还有烦死了