題目描述

Farmer John‘s NNN 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…61 \ldots 61…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.

給你n個數，求一個最長的區間，使得區間和能被7整除

輸入輸出格式

輸入格式：

The first line of input contains NNN (1≤N≤50,0001 \leq N \leq 50,0001≤N≤50,000 ). The next NNN

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

0…1,000,0000 \ldots 1,000,0000…1,000,000 ).

輸出格式：

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.

輸入輸出樣例

7
3
5
1
6
2
14
10

5

說明

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

同余的性質

(a mod m) + (b mod m) ≡ a + b (mod m)
(a mod m) - (b mod m) ≡ a - b (mod m)
(a mod m) * (b mod m) ≡ a * b (mod m)

給出序列，求出最長的區間，滿足區間和為7的倍數
預處理出前綴和
(pre[j] - pre[i ]) mod 7 = 0
=> pre[i] ≡ pre[j] (mod 7)
維護7個值，分別表示模7余0到6的最前d的前綴和
掃一遍過去就行了

 1 //2018年2月14日14:15:25
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 const int N = 100001;
 8 int n, a[N], sum[N], last[N], first[N], ans;
 9 
10 int main(){
11     scanf("%d", &n);
12     for(int i=1;i<=n;i++){
13         scanf("%d", &a[i]);
14         sum[i] = (sum[i-1] + a[i]) % 7;
15     }
16     for(int i=1;i<=n;i++)
17         last[sum[i]] = i;
18     for(int i=n;i>=1;i--)
19         first[sum[i]] = i;
20     for(int i=0;i<7;i++)
21         ans = max(ans, last[i]-first[i]);
22     printf("%d\n", ans);
23 
24     return 0;
25 }

[USACO16JAN]子共七Subsequences Summing to Sevens