hdu1003 Max Sum（经典dp ）

 

A - 最大子段和

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

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

 

 

题解：

求最大子序列和，及其初始和结束位置。  如果有多解，输出最先算出的。

两种方法，一看就懂得了

 

#include <cstdio>
int main()
{
    int T,k, R, L, Max, sum, t, total=, n;
    scanf("%d", &T);
    while(T--)
    {
        sum = Max =-;  //足够小就好
        scanf("%d", &n);
        for(int i=; i<=n; i++)
        {
            scanf("%d", &t);
            if (sum<)
            {
                sum = t, k = i;
            }
            else
                sum += t;
            if (Max < sum) Max = sum, L = k, R = i;  //L R分别为初始和结束位置
        }
        printf("Case %d:\n", total++);
        printf("%d %d %d\n", Max, L, R );
        if (T) printf("\n");
    }
    return ;
}

#include<iostream>
using namespace std;
int a[],b[];
int main()
{
    int n,T,s,t,max,total,k=;
    cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i=; i<=n; i++)
            cin>>a[i];
        b[]=a[];
        for(int i=; i<=n; i++)
        {
            if(b[i-]<)
                b[i]=a[i];
            else
                b[i]=b[i-]+a[i];
        }
        max=b[];
        s=;
        for(int i=; i<=n; i++)
        {
            if(b[i]>max)
            {
                max=b[i];
                s=i;
            }
        }
        t=s;
        total=;
        for(int i=s; i>=; i--)
        {
            total+=a[i];
            if(total==max) t=i;
        }
        cout<<"Case "<<k++<<":"<<endl;
        cout<<max<<" "<<t<<" "<<s<<endl;
        if(T) cout<<endl;
    }
}