HDU 1003 Max Sum (入门DP[1] 最大连续子序列)

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aozil_yang

最新推荐文章于 2025-08-06 19:08:55 发布

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分类专栏： HDU DP 文章标签： c语言 dp

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本文介绍了一个经典的编程问题——MaxSum问题，旨在寻找整数数组中和最大的连续子序列及其起始和终止位置。通过动态规划的方法，文章给出了详细的算法实现步骤及代码示例。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 210999 Accepted Submission(s): 49522

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

Author

大体题意：

给你一个整数数组，求的一个连续子序列，使得连续子序列和最大，并且输出子序列的和和起始位置和终止位置！

思路：

据教练指导，入门DP 的第一题！

设计：

dp[i] 表示1.2.3....i 中最大值。

那么转移方程：

当dp[i-1] < 0时，那么dp[i]最优解肯定是a[i]了，

否则就是dp[i-1] + a[i]:

然后遍历一遍求得最大值并且找到最右端位置，然后在从最右端向左找找到最左端

注意：

找最左端时，不要发现一个最左端就break 要等他找完，因为可能不是最长！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100000 + 10;
int dp[maxn],a[maxn];
int main(){
    int T,cnt = 0;
    scanf("%d",&T);
    while(T--){
        memset(dp,0,sizeof dp);
        int n;
        scanf("%d",&n);
        for (int i = 1; i <= n; ++i)scanf("%d",&a[i]);
        dp[1]=a[1];
        for (int i = 2; i <= n; ++i){
            if (dp[i-1] < 0)dp[i] = a[i];
            else dp[i] = dp[i-1] + a[i];
        }
        int tmp_max = dp[1],r = 1;// r 一定要初始化为1，可能全是负数！！
        for (int i = 2; i <= n; ++i){
            if (dp[i] > tmp_max){
                tmp_max = dp[i];
                r = i;
            }
        }
        int t = 0,l;
        for (int i = r; i >= 1; --i){
            t += a[i];
            if (t == tmp_max)l = i;
        }
        if (cnt++)printf("\n");
        printf("Case %d:\n",cnt);
        printf("%d %d %d\n",tmp_max,l,r);
    }
    return 0;
}