【HDU 杭电 1003 Max Sum】

Max Sum

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).

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<queue>#include<stack>#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define  K 100011#define INF 0x3f3f3fusing namespace std;int  pa[K];int  st[K];//当前的最大连续子序列的最大和 int main(){   int i,j;   int T;   int n;   int ans;//最大连续子序列的和    int a,b;//子序列的首尾位置    int test=1;   int p=0;//最大连续子序列和的尾位置    scanf("%d",&T);   while(T--)   {    if(p)    printf("\n");    p=1;    scanf("%d",&n);    for(i=1;i<=n;i++)    scanf("%d",&pa[i]);    st[1]=pa[1];ans=pa[1];    b=1;     for(i=2;i<=n;i++)    {        st[i]=max(pa[i],pa[i]+st[i-1]); //当前连续子序列的最大和         if(st[i]>ans)//更新最大和         {            ans=st[i];            b=i;        }    }    a=b;    for(i=b;i>=1;i--)//求首位置     {        if(st[i]>=0)        a=i;        else        break;    }    printf("Case %d:\n",test++);    printf("%d %d %d\n",ans,a,b);   }   return 0;}