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;

    }

}

巴特西

hdu1003 Max Sum（经典dp ）

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