D - 最大连续和 Easy

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

题意是要求最大连续和，与以往不同的是追加输出子串的头尾位置，这个可以用结构体记录，最坑的是没有注意到格式说每两个case间要空一行。

代码：

#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int  a[100005];

struct studen
{
        long long sum;
        int l,r;
}dp[100005];

long long maxw(long long  x,long long  y)
{
        if(x>y)
                return x;
        else
                return y;
}
int main()
{
    int n,i,m,s=0;
    scanf("%d",&m);
    while(m--)
    {
            s++;
           scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
            scanf("%d",&a[i]);
            dp[i].sum=a[i];dp[i].l=i;dp[i].r=i;
    }
   studen max1=dp[1]; 

     for(i=2;i<=n;i++)
    {
         dp[i].sum=maxw(a[i],dp[i-1].sum+a[i]);
        if(dp[i-1].sum>=0)
                dp[i].l=dp[i-1].l;
        else
                  dp[i].l=i;
         dp[i].r=i;
          if(dp[i].sum>=max1.sum)
         {
                 max1=dp[i];

         }
    }
   
    printf("Case %d:\n",s);
    printf("%lld %d %d\n",max1.sum,max1.l,max1.r);
     if(m)
        printf("\n");


    }

    return 0;
}

D - 最大连续和 Easy

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