HDU 1003 Max Sum【最大连续和】

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

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4
Case 2:
7 1 6

code：

#include<stdio.h>
int main()
{
 int p,t,s,sum,tot,e,i,n,k=0,st;
 scanf("%d",&t);
 while(t--)
 {
  tot=0;
  st=e=1;
  sum=-9999;
  if(k++)
   putchar('\n');
  scanf("%d",&n);
   for(i=1;i<=n;i++)
   {
    scanf("%d",&p);
    tot+=p;
    if(tot>sum)
    {
     sum=tot;
     s=st;
     e=i;
    }
    if(tot<0)
    {
     tot=0;
     st=i+1;
    }
   }
  printf("Case %d:\n%d %d %d\n",k,sum,s,e);
 }
 return 0;
}

 附：分治法求最大和

code：

#include<stdio.h>
int s,e;
int maxsum(int* a,int x,int y)         //返回数组左闭右开区间[x,y)中的最大连续和
{
    int i,m,v,l,r,max;
    if(y-x==1)
        return a[x];                  //只有一个元素，直接返回
    m=x+(y-x)/2;                      //分治第一步：划分成[x,m)和[m,y)
    max=maxsum(a,x,m)>maxsum(a,m,y)?maxsum(a,x,m):maxsum(a,m,y);//分治第二步：递归求解
    v=0;
    l=a[m-1];                 //分治第三步：合并（1）—从分界点开始往左的最大连续和L
    for(i=m-1;i>=x;i--)
    {
        v+=a[i];
        l=l>v?l:v;
    }
    v=0;
    r=a[m];                  //分治第三步：合并（2）—从分解点开始往右的最大连续和R
    for(i=m;i<y;i++)
    {
        v+=a[i];
        r=r>v?r:v;
    }
    return max=max>(l+r)?max:(l+r);//把子问题的解与l和r比较
}
int main()
{
    int n,i,k;
    int a[1000];
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        k=maxsum(a,0,n);
        printf("%d\n",k);
    }
    return 0;
}