hdu Max Sum

Max Sum

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

 

代码：

#include<stdio.h>#include<string.h>#define max(a,b)  (a>b?a:b)#define min(a,b)  (a<b?a:b)#define N  100000+10int a[N];int main(){    int t,m=0;    scanf("%d",&t);    while(++m<=t)    {        int n,i,j;        scanf("%d",&n);        for(i=1; i<=n; i++)            scanf("%d",&a[i]);        int maxn=-10000,x1=1,cnt=0,x,y;        for(i=1; i<=n; i++)        {            cnt+=a[i];            if(cnt>maxn)                x=x1,y=i,maxn=cnt;            if(cnt<0)                cnt=0,x1=i+1;        }        printf("Case %d:\n%d %d %d\n",m,maxn,x,y);        if(m!=t)            printf("\n");    }    return 0;}

ps：我还以为 5    1  0 0 0 0    这组数据的答案是1  1 5呢，WA了n次，也没谁了