HDU 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 86447    Accepted Submission(s): 20035

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

 

分析：求最大的连续子列和，求最大连续的子列的最大和
算法：1.一直相加直到负数，然后归零，再相加的期间会有一个最大的和，记录 
 2.重复第一步，直到输入完毕。
证明如下：
  原理：如果在连续子串 <a1,a2,a3,...,an> 中Sn<0;S1,S2,...S(n-1)均大于或等于0（Sn为前n个数的和）
那么肯定存在一个最大的Si,Si也就是a1到an这段连续子串的最大和。
我们知道a1>0时任意的 <ai...an> (i>1) 的和都小于0，（Sn-S1，Sn-s2，。。。。<0）
所以这些元素就不能与an后的元素构成能获得最大和的子串(如果a1=0则a2>=0,情况类似);

同时在 <ai...aj> (i>1,j<=n) 也不能或得比Si大的和。（Sn-Si-Sj<0）

代码：

#include<stdio.h>#define MAX 100020int a[MAX];int main(){    int T;    scanf("%d",&T);    int count=0;    while(T--)    {        count++;        int M,i,j,start,end;        long max=-3000,sum=0;        scanf("%d",&M);        for(i=0,j=0;i<M;i++)        {            scanf("%d",&a[i]);            sum+=a[i];            if(sum>max)            {                max=sum;                start=j;                end=i;            }            if(sum<0)            {                sum=0;                j=i+1;            }        }        printf("Case %d:\n",count);        printf("%ld %d %d\n",max,start+1,end+1);        if(T!=0)            printf("\n");    }    return 0;}

注意：一定要先判断sum>max的情况，因为可能会出现全是负数的情况，若先判断sum<0的话，则sum会一直是0