HDU 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 247744    Accepted Submission(s): 58531

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

 

分析：经典的最大字段和+记录路径理解了转换公式就ok了；

详见代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){    int n,t;    int a[100005];    int sum,MAX;    int start,end,tem;    int index=0;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);         MAX=-100000000;        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        sum=start=tem=0;        for(int i=0;i<n;i++)        {            if(sum>0)//可能有的同学会纠结这个代码连实示例都不过，改为sum>=0就行了，其实两种方法都能找到最优解，只是区间记录的不一样                 sum+=a[i];            else            {                tem=i;                sum=a[i];            }            if(sum>MAX)            {                MAX=sum;                start=tem;//这里必须借助tem变量，记录随时变化的左区间                 end=i;            }        }        printf("Case %d:\n",++index);        printf("%d %d %d\n",MAX,start+1,end+1);        if(t>0)            printf("\n");    }    return 0;}