HDU1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 194034 Accepted Submission(s): 45254

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
说实话是看了别人的思路过的。基本思路就是一个一个加起来找出最大值，当累加的值为负数时，就归为零，从下一个继续开始。
#include<stdio.h>#define INF 0x3f3f3f3fint main(){    int t,n,a,i,j,sum,max,l,r,cas=0;    scanf("%d",&t);    while(t--)    {        cas++;        sum=j=l=r=0;        max=-INF;        scanf("%d",&n);        for(i=1; i<=n; i++)        {            scanf("%d",&a);            sum+=a;            if(sum>max)//比较当前sum与之前最大的连续子序列和，一旦比之前的max更大，就进行替换。            {                max=sum;                l=j+1;                r=i;            }            if(sum<0)//sum为负数，就归为零，重新开始。            {                sum=0;                j=i;//重新开始需将j进行更改。            }        }        printf("Case %d:\n",cas);        printf("%d %d %d\n",max,l,r);        if(t)            printf("\n");    }    return 0;}