[USACO16JAN] Subsequences Summing to Sevens S

题目描述

Farmer John's N cows are standing in a row, as they have a tendency to do from time to time.  Each cow is labeled with a distinct integer ID number so FJ can tell them apart. FJ would like to take a photo of a contiguous group of cows but, due to a traumatic  childhood incident involving the numbers 1 ... 6, he only wants to take a picture of a group of cows if their IDs add up to a multiple of 7.

Please help FJ determine the size of the largest group he can photograph.

给你n个数，分别是a[1],a[2],...,a[n]。求一个最长的区间[x,y]，使得区间中的数(a[x],a[x+1],a[x+2],...,a[y-1],a[y])的和能被7整除。输出区间长度。若没有符合要求的区间，输出0。

输入格式

The first line of input contains N (1 <= N <= 50,000).  The next N

lines each contain the N integer IDs of the cows (all are in the range

0 ... 1,000,000).

输出格式

Please output the number of cows in the largest consecutive group whose IDs sum

to a multiple of 7.  If no such group exists, output 0.

样例 #1

样例输入 #1
7
3
5
1
6
2
14
10

样例输出 #1
5

提示

In this example, 5+1+6+2+14 = 28.

代码：

#include 
#include 
using namespace std;

const int N = 100010; // 数组的最大长度

int pre[N]; // 存储前缀和的数组
int first[7], last[7]; // 存储前缀和首次出现位置和最后一次出现位置
int mx = -1; // 用于记录最大长度

int main()
{
    int n; // 数组长度
    cin >> n;

    // 计算前缀和
    for (int i = 1; i <= n; i++)
    {
        cin >> pre[i];
        pre[i] = (pre[i] + pre[i - 1]) % 7; // 计算前缀和并取模
    }

    // 计算前缀和首次出现位置
    for (int i = n; i > 0; i--)
        first[pre[i]] = i;

    first[0] = 0; // 处理前缀和为 0 的情况，从数组开头开始的子数组

    // 计算前缀和最后一次出现位置
    for (int i = 1; i <= n; i++)
        last[pre[i]] = i;

    // 遍历所有前缀和，计算最大子数组长度
    for (int i = 0; i <= 6; i++)
        mx = max(last[i] - first[i], mx);

    // 输出结果，即满足和为 7 的倍数的最大子数组长度
    cout << mx;

    return 0;
}

这段代码的主要思想是通过计算前缀和，然后使用 first 和 last 数组来记录每个前缀和首次出现位置和最后一次出现位置。通过遍历所有前缀和，计算每个前缀和对应的最大子数组长度，找到最大值，最终输出满足和为 7 的倍数的最大子数组长度。这个算法的时间复杂度是 O(n)。