杭电ACM1003



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

算法分析：求最大字段和，d[i]表示已i 结尾（字段和中包含 i ）在 a[1..i] 上的最大和，d[i]=(d[i-1]+a[i]>a[i])?d[i-1]+a[i]:a[i];

max = {d[i],1<=i<=n} ;

#include<iostream>
#define N 100010
using namespace std;
int a[N],d[N];
int main()
{
    int test,n,i,max,k,f,e;
    cin>>test;
    k=1;
    while(test--)
    {
        cin>>n;
        for(i=1;i<=n;i++)
            cin>>a[i];
        d[1]=a[1];
        for(i=2;i<=n;i++)
        {
            if(d[i-1]<0) d[i]=a[i];
            else d[i]=d[i-1]+a[i];
        }
        max=d[1];e=1;
        for(i=2;i<=n;i++)
        {
            if(max<d[i])
            {
                max=d[i];e=i;
            }
        }
        int t=0;
        f=e;
        for(i=e;i>0;i--)
        {
            t=t+a[i];
            if(t==max)    f=i;
        }
        cout<<"Case "<<k++<<":"<<endl<<max<<" "<<f<<" "<<e<<endl;
        if(test) cout<<endl;
    }
    return 0;
}

只处理最大和不处理位置

#include<cstdio>
int main()
{
    int n,test,ans,t,a,i;
    scanf("%d",&test);
    while(test--)
    {
        scanf("%d",&n);
        scanf("%d",&a);
        ans=t=a;
        for(i=1;i<n;i++)
        {
            scanf("%d",&a);
            if(t<0) t=a;
            else t=t+a;
            if(ans<t) ans=t;
        }
        printf("%d\n",ans);
    }
    return 0;
}