UC知道ACM MAX SUM
来源：百度知道 编辑：UC知道 时间：2024/05/13 04:44:10
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
InputThe first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
OutputFor each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
我自己编的但总是运行出错，请高手矫正下
#include <std
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
InputThe first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
OutputFor each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
我自己编的但总是运行出错，请高手矫正下
#include <std
z=sum=s=y=0;
z设为0，如果输入的数值全为负数，可以吗？
好像算法不对，感觉用DP解吧。
而且你的程序好像有一组很简单的测试数据都过不了！
4 1 2 -5 1
结果应该是3吧？（只是粗略的看了一下），而你的程序输出来应该不是3！
你试试吧，这种类型用贪心应该也可以！（仅供参考）