MAX SUM

先粘题目

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 167166    Accepted Submission(s): 39004

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.

 

Input

The 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).

 

Output

For 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 Input

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1: 14 1 4 Case 2: 7 1 6

 

 

很明显是最长子序列问题。一开始我也是就这么写的。但是处理起来又貌似有点不一样。数组太大。总是Runtime Error (ACCESS_VIOLATION)。我于是百度了一下。借鉴别人借鉴的别人的。

 

http://blog.csdn.net/akof1314/article/details/4757021

 

 

自己Ac代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<climits>

using namespace std;

#define N 11000

int a[N];

int main()
{
    int t, i, k, n, max, x, temp, p1, p2, now;

    scanf("%d", &t);

    for(k = 1; k <= t; k++)
    {
        scanf("%d%d", &n, &temp);
        
        x = p1 = p2 = 1;
        max  = now = temp;//初始化
        
        for(i = 2; i <= n; i++)
        {
            scanf("%d", &temp);

            if(now+temp < temp)//判断这个数前边的数是加还是不加，如果不加就把起始位置置为i
            {
                now = temp;
                x = i;
            }
            else now += temp;

            if(now > max)//记录当前最大值时的起始及结束位置坐标
            {
                p1 = x;
                p2 = i;
                max = now;
            }
        }


        if(k != 1)
            printf(" ");//格式控制，也可以由k是否等于n判断输出之后是否输出换行
        printf("Case %d: ", k);
        printf("%d %d %d ", max, p1, p2);
    }
    return 0;
}

IDEAS：

计算机，或者程序自身是有记忆功能的。当你处理每一步时，你都可以利用前边已经计算过的值或者结果处理你下一步如何操作。尤其是当数组很大，或者时间超限的时候。既可以减少运行内存还能降低运行时间。。你当然也可以选择继续没想法下去