当前你的浏览器版本过低,网站已在兼容模式下运行,兼容模式仅提供最小功能支持,网站样式可能显示不正常。
请尽快升级浏览器以体验网站在线编辑、在线运行等功能。

建议使用的浏览器:

谷歌Chrome 火狐Firefox Opera浏览器 微软Edge浏览器 QQ浏览器 360浏览器 傲游浏览器

1858:Max Partial Value I

题目描述
HenryFour has a number of stones which have different values from -4444 to 4444. He puts N stones in a line and wants to find the max partial value of these N stones.

Assume the values of the N stones in line are: v1, v2, v3, v4, ..., vN. The partial vaule of stones from Lth stone to Rth stone (1 ≤ L ≤ R ≤ N) is the sum of all the stones between them. i.e. PartialV(L, R) = v[L] + v[L+1] + .... + v[R] (1 ≤ L ≤ R ≤ N)

Since the number of stones (N) is very very large, it is quite difficult for HenryFour to find the max partial value. So could you develop a programme to find out the answer for him?
输入解释
There are several test cases in the input data. The first line contains a positive integer T (1 ≤ T ≤ 14), specifying the number ot test cases. Then there are T lines. Each of these T lines contains a positive number N followed by N integers which indicate the values of the N stones in line.
1 ≤ N ≤ 1,000,000
-4444 ≤ v[i] ≤ 4444
输出解释
Your program is to write to standard output. For each test case, print one line with three numbers seperated by one blank: P L R. P is the max partial value of the N stones in line. L and R indicate the position of the partial stones. If there are several Ls and Rs that have the same value PartialV(Li, Ri) = P, please output the minimum pair. For pair (Li, Ri) and (Lj, Rj), we define (Li, Ri) < (Lj, Rj) if and only if: Li < Lj or (Li == Lj and Ri < Rj)
输入样例
3
4 32 -39 -30 -28
8 1 2 3 -10 1 -1 5 1
10 14 -12 -8 -13 3 5 42 -24 -32 -12
输出样例
32 1 1
6 1 3
50 5 7

提示
 Huge input and output,scanf and printf are recommended.
来自杭电HDUOJ的附加信息
Author HenryFour@TJU
Recommend lcy

该题目是Virtual Judge题目,来自 杭电HDUOJ

源链接: HDU-1858

最后修改于 2020-10-25T22:48:21+00:00 由爬虫自动更新

共提交 0

通过率 --%
时间上限 内存上限
1000/5000MS(Java/Others) 32768/65535K(Java/Others)