HDU1003 Max Sum

                                                                 Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 112106    Accepted Submission(s): 25891

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


解题思路：求和最大连续子串。。。。。。。。。用max记录当前最大和，一旦sum小于0，则置0从新累加。

#include<stdio.h>int main(){    int a[100002];    int t,n;    int sum;    int max1,maxx,maxy;    int i,j,k;    scanf("%d",&t);    for(k=1;k<=t;k++)    {        max1=-999999999;        sum=0;        scanf("%d",&n);        for(i=0;i<n;i++)            scanf("%d",&a[i]);        j=0;        for(i=0;i<n;i++)        {            //printf("\ni=%d",i);            sum+=a[i];            // printf("%d ",sum);            if(sum>max1)            {                max1=sum;                maxx=j+1;                maxy=i+1;                //printf("-------%d %d %d\n",max1,maxx,maxy);            }            if(sum<0)            {                sum=0;                j=i+1;            }        }        if(k!=1)            printf("\n");        printf("Case %d:\n",k);        printf("%d %d %d\n",max1,maxx,maxy);    }    return 0;}