hdu1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98850    Accepted Submission(s): 22737

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

 

贪心算法：

#include<stdio.h>
int main()
{
    int m; //case数
    int n; // 数列长度
    int i,j;
    int p,start,temp,end;
    int max,sum;
    scanf("%d",&m);
    for(i = 1; i <= m; i++)
    {
        sum = 0;
        temp = 1;            // 扫描到的第一个子段肯定是起始位置为1
        max = -1001;
        scanf("%d",&n);
        for(j = 1; j <= n; j++)
        {
            scanf("%d",&p);
            sum += p;
            if(sum > max)
            {
                max = sum;      // 遇到更大的子段和，更新起始，结束位置
                start = temp;
                end = j;
            }
            if(sum < 0)
            {
                sum = 0;       //当前子段和归零
                temp = j + 1;   //下一位置为起始位
            }
        }
        printf("Case %d:\n%d %d %d\n",i,max,start,end);
        if(i < m) printf("\n");
    }
    return 0;
}

动态规划

#include<cstdio>
using namespace std;
int dp[100010]; // dp[i]存储以第i个元素为结尾的最大子段和
int main()
{
    
    int T;
    scanf("%d", &T);
    for(int I = 1;I <= T; I++)
    {
        int n,start = 1,end = 1,ans, temp = 1;

        scanf("%d",&n);
        for(int i = 1;i <= n;i ++)
            scanf("%d",dp + i);

        ans = dp[1];
        for(int i = 2;i <= n;i ++)
        {
            int t = dp[i] + dp[i - 1];

            if(t >= dp[i])	//如果加上第i个元素比之前的和大，						

                dp[i] = t;	//那么第i个元素为结尾的最大子段和	
            else		//就是第i个元素和前一个元素为结尾的最大子段和的和
             	temp = i;

            if(ans < dp[i])	//如果以第i个元素为结尾的最大子段和比当前最大的还大，就更新
            {
                ans = dp[i];
                start = temp;
                end = i;
            }
        }

        printf("Case %d:\n",I);
        printf("%d %d %d\n",ans,start,end);
        if(I < T)printf("\n");
    }
    return 0;
}

通过滚动数组减少内存的动态规划

#include<cstdio>
using namespace std;
int main()
{
     int T, dp[2];		//因为再上一个代码里面dp[i]只是和前一个元素dp[i - 1]有关系，
    scanf("%d", &T);		//所以没必要开那么大的数组，每次用完i-1个元素就可以覆盖它
    for(int I = 1;I <= T; I++)
    {
        int n, start = 1, end = 1, ans, temp = 1;

        scanf("%d",&n);
        scanf("%d",&ans);
        dp[0] = ans;
        for(int i = 2,j = 1;i <= n;i ++)
        {
            scanf("%d",dp + j);
            int t = dp[j] + dp[j ^ 1];

            if(t >= dp[j])
                dp[j] = t;
            else
             temp = i;

            if(ans < dp[j])
            {
                ans = dp[j];
                start = temp;
                end = i;
            }
           j ^= 1;
        }

        printf("Case %d:\n",I);
        printf("%d %d %d\n",ans,start,end);
        if(I < T)printf("\n");
    }
    return 0;
}