带运算符号的最大子序列

带运算符号的最大子序列

目录

问题大意：

样例输入：

样例输出：

解决方案：

问题描述：

 题目：

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.

问题大意：

该问题输入一个数n,接下来有n行数据，第一个数输入m,后面有m个数据，m个数据组成序列，求一个子序列所得的最大值，并且求出子序列的开始位置（头），结束位置（尾）

样例输入：

Sample Input

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

样例输出：

Sample Output

Case 1：

14 1 4 

Case 2：

7 1 6

解决方案：

1.        

   #include "stdio.h"
   #include "string.h"
   #include "algorithm"
   using namespace std;
   int a[101000];
   int main ()
   {int t,i,k,n,j=0;int s,g;scanf("%d",&t);while(t--){j++;int intf=-999999;int sum=0;memset(a,0,sizeof(a));scanf("%d",&n);for(i=1; i<=n; i++)scanf("%d",&a[i]);int head=1;//记录头部 for(i=1; i<=n; i++){sum=sum+a[i];// 依次求和 //如果和大于0，且再向后加数的时候比前一个和大 ;   //更新结束位置（尾部）；  if(sum>intf)   {intf=sum;     s=head;g=i;//记录尾部； }     if(sum<0){sum=0;head=i+1;}// 和小于0// 更新头部； }printf("Case %d:n",j);printf("%d %d %dn",intf,s,g);if(t) printf("n");}return 0;
   }