训练3 习题1

题目：

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.<br>

 

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).<br>

 

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.<br>

 

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

思路：每输入一个数   进行判断  如果前项小于0  直接改变s      其他before直接相加  和max 比较 相应处理

如：

6 -1 5 4 -7
当第一个时  初始化 
第二个 前项before 是正的 before加上这个数  之后和max比较 不改变max 
第三步 前项before 是正的 before加上这个数  之后和max比较 改变max 和e
第四个 前项before 是正的 before加上这个数  之后和max比较 不改变max 
之后  一样

#include<stdio.h>

int main()
{
    int i,ca=1,t,s,e,n,x,now,before,max;
    scanf("%d",&t);
    while(t--)
    {
       scanf("%d",&n);
       for(i=1;i<=n;i++)
       {
         scanf("%d",&now);
         if(i==1)
         {
            max=before=now;
            x=s=e=1;
         }
         else {
             if(now>now+before)
             {
                before=now;
                x=i;
             }
             else before+=now;
              }
         if(before>max)
           max=before,s=x,e=i;
       }
       printf("Case %d:\n%d %d %d\n",ca++,max,s,e);
       if(t)printf("\n");
    }
    return 0;
}