hdu 1003 Max Sum【一维数组最大连续和】

链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1003

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28508#problem/A

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 113473    Accepted Submission(s): 26241

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

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

 

Author

Ignatius.L

 

算法：做的时候感觉是贪心，本质应该也算是DP了

思路：

      对于数字：a[0]----a[n-1]
     状态转移方程：dp[0] = a[0]
                      dp[i] = max(dp[i-1], 0)+a[i]
                      最后遍历 dp[] 找出最大的 dp[i] 就是这一行数的最大连续和
     状态转移方程样例分析：
       a[]： 6 -1 3 -10 8 7
       dp[]：6  5 8  0  8 15

     关于下标,判断dp[i-1]是不是 0 的时候更新就可以了,这么写主要是为了方便下一题的描述

    事实上直接  a[0] = a[0]

                      a[i] = max(a[i-1], 0) + a[i]  更新时注意下下标的处理就可以了，最后再输出最大的 a[i] 

code：

#include<stdio.h>int main(){    int T;    int n;    int a; //存输入的数    scanf("%d", &T);    for(int kcase = 1; kcase <= T; kcase++)    {        scanf("%d", &n);        scanf("%d", &a);        int sum = a; //最大值        int Max = a; //中间比较变量        int index1 = 1; //起点        int index2 = 1; //终点        int st = 1; //中间比较变量的起点        if(Max < 0) //注意        {            Max = 0;            st = 2;        }        for(int i = 2; i <= n; i++)        {            scanf("%d", &a);            Max += a;            if(Max > sum)            {                sum = Max;                index1 = st;                index2 = i;            }            if(Max < 0) //注意别加 else 否则上面判断后,不会进入下面, 开始没有注意,坑了很久            {                Max = 0;                st = i+1;            }        }        printf("Case %d:\n", kcase);        printf("%d %d %d\n", sum, index1, index2);        if(kcase != T) printf("\n"); //注意格式    }    return 0;}