FZU2013 A short problem

 Problem Description  题目链接

The description of this problem is very short. Now give you a string(length N), and ask you the max sum of the substring which the length can't small than M.

 Input

The first line is one integer T(T≤20) indicates the number of the test cases. Then for every case, the first line is two integer N(1≤N≤1000000) and M(1≤M≤N).

Then one line contains N integer indicate the number. All the number is between -10000 and 10000.

 Output

Output one line with an integer.

 Sample Input

2

5

1 1 -2 -2 -2 1

5

2 1 -2 -2 -2 1

 Sample Output

1

-1

#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAX = 1000005;

/*
    思路：最长递增子序列，只不过有个数限制至少m个，所以转移方程有点变化
    sum[i]表示前缀和 dp[i]表示i所在的最长序列和
    转移方程为 dp[i]=max(dp[i-1]+nums[i],sum[i]-sum[i-m])
    后面还有一个加速版
*/

int nums[MAX];
LL sum[MAX];
LL dp[MAX];

int main()
{
    int n, t, m;
    scanf("%d", &t);
    sum[0] = 0;
    while (t--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &nums[i]);
            sum[i] = sum[i - 1] + nums[i];
        }
        LL maxNum = sum[m];
        dp[m] = maxNum;
        for (int i = m + 1; i <= n; i++)
        {
            dp[i] = max(dp[i - 1] + nums[i], sum[i] - sum[i - m]);
            if (maxNum < dp[i])
                maxNum = dp[i];
        }
        printf("%I64d
", maxNum);
    }
}

省内存版：

#include <iostream>
#include <algorithm>
using namespace std;

/*
    迭代变量缩小内存
*/

typedef long long LL;

int nums[1000005];

int main()
{
    int n, t, m;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &m);
        LL lastSum = 0, preSum = 0;
        for (int i = 1; i <= m; i++)
        {
            scanf("%d", &nums[i]);
            lastSum += nums[i];
        }
        LL maxNum = lastSum, temp = lastSum;
        for (int i = m + 1; i <= n; i++)
        {
            scanf("%d", &nums[i]);
            lastSum += nums[i];
            preSum += nums[i - m];
            temp = max(temp + nums[i], lastSum - preSum);
            if (maxNum < temp)
                maxNum = temp;
        }
        printf("%I64d
", maxNum);
    }
}