最大连续子序列&&MAX SUM

//错的莫名其妙的O w O

第二个的格式也是莫名其妙的

 

Input
测试输入包含若干测试用例，每个测试用例占2行，第1行给出正整数K( < 10000 )，第2行给出K个整数，中间用空格分隔。当K为0时，输入结束，该用例不被处理。

Output
对每个测试用例，在1行里输出最大和、最大连续子序列的第一个和最后一个元
素，中间用空格分隔。如果最大连续子序列不唯一，则输出序号i和j最小的那个（如输入样例的第2、3组）。若所有K个元素都是负数，则定义其最大和为0，输出整个序列的首尾元素。

Sample Input
6
-2 11 -4 13 -5 -2
10
-10 1 2 3 4 -5 -23 3 7 -21
6
5 -8 3 2 5 0
1
10
3
-1 -5 -2
3
-1 0 -2
0

Sample Output
20 11 13
10 1 4
10 3 5
10 10 10
0 -1 -2
0 0 0

 

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

 

#include <stdio.h>
#include <string.h>
int main()
{
      int first,last,temp,n,i,j,flag,thissum;
      int a[22222];
      while(scanf("%d",&n)&&n)
      {
          flag=0;
          thissum=0;
          memset(a,0,sizeof(a));
          for(i=1;i<=n;i++)
          {
                scanf("%d",&a[i]);
                if(a[i]>=0)
                    flag=1;
          }
                 first=last=temp=1;
          if(!flag)
          {
              printf("0 %d %d\n",a[1],a[n]);
              continue;
          }
          int max=-33333;
          for(i=1;i<=n;i++)
          {
              thissum+=a[i];
                if(thissum>max)
                {
                    max=thissum;
                    first=temp;
                    last=i;
                }
                if(thissum<0)
                {
                    thissum=0;
                    temp=i+1;
                }
          }
          printf("%d %d %d\n",max,a[first],a[last]);
      }
      return 0;
}