Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 255311    Accepted Submission(s): 60666

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).

    题解：

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
      
       using namespace std;int main(){    int N,n,a,first,last; cin>>N; for(int i=1;i<=N;i++) {     cin>>n;     int max=-1000,sum=0,k=1;     for(int j=1;j<=n;j++)    {        cin>>a;         sum=sum+a;         if(sum>max)//如果当前的最大值大于以前的最大值  更新 sum 与max 都是一种计算的结果和属性相同         {         first=k;         last=j;         max=sum;         }         if(sum<0)         //else         {             k=j+1;             sum=0;         }     }     cout<<"Case "<
       <<":"<
      

#include 
      
       int main(){    int z,n,max,sum;    int a,b,A,B,t;    scanf("%d",&z);    for(int k=1;k<=z;k++)    {        scanf("%d",&n);        sum = max = -1001;        for(int i=1;i<=n;i++)        {            scanf("%d",&t);            if(sum+t < t)                sum = t , a = b = i;      //a、b记录当前连续子序列的起始、结束位置            else                sum += t , ++b;            if(max < sum)//附一篇另一种代码，在这儿不知道这个if语句的作用                max = sum , A = a , B = b;        }        printf("Case %d:\n%d %d %d\n",k,max,A,B);        if(k-z) puts("");    }	return 0;}