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HDU 1003 Max Sum

编程语言 m0_37253730 19℃ 0评论

简单dp入门


其实感觉这题贪心可写,分治可写

题目:


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

思路:简单的dp,然而我开的数组明显比别人多一个,甚至有些人没开数组。。


我的想法是先开一个数组d[maxn] 保存以i为结尾的最大的连续和的值,而sti[i]保存这些值以哪一个下标为开始。


temp 是当前下标为i的值


故有d[i] = max(temp, temp + d[i-1]); 当temp值比较大时,sti[i] = i,另外一种情况时, sti[i] = sti[i-1]

代码:

#include 
#include 
#include 
#include 
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f

int d[maxn];
int sti[maxn];
int main()
{
    //freopen("D://in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    for(int kase = 1; kase <= t; kase ++)
    {
        int n;
        scanf("%d",&n);
        int temp;
        scanf("%d",&temp);
        d[0] = temp;
        for(int i = 1; i < n ; i++)
        {
            scanf("%d",&temp);
            d[i] = max(d[i-1] + temp, temp);
            if(d[i-1] + temp <= temp)
                sti[i] = i;
            else
                sti[i] = sti[i-1];
        }

        int ans = -inf;
        int ansj = -1;
        int ansi = -1;
        for(int i = 0; i < n ; i++)
        {
            if(ans < d[i])
            {
                ans = d[i];
                ansj = i;
                ansi = sti[i];
            }
        }
        if(kase != 1)
            printf("\n");
        printf("Case %d:\n",kase);
        printf("%d %d %d\n",ans,ansi+1,ansj+1);
    }

    return 0;
}

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