BZOJ 4511: [Usaco2016 Jan]Subsequences Summing to Sevens

4511: [Usaco2016 Jan]Subsequences Summing to Sevens

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 197  Solved: 149
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Description

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.

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

).

给定长度为N的数列A，求A最长的连续子序列，满足子序列中元素的和是7的倍数。输出最大的长度。如果不存在满

足条件的子序列，输出0。1 ≤ N ≤ 50000, 0 ≤ A_i ≤ 10 ^ 6

Input

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.

Output

You may want to note that the sum of the IDs of a large group of cows might be too large to fit into a standard 32-bit integer. If you are summing up large groups of IDs, you may therefore want to use a larger integer data type, like a 64-bit "long long" in C/C++.

Sample Input

7
3
5
1
6
2
14
10

Sample Output

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

刷水有益健康

维护前缀和就行了

BJ处理了一下连续0的情况

搞的rank超低

#include<cmath>#include<ctime>#include<cstdio>#include<cstring>#include<cstdlib>#include<complex>#include<iostream>#include<algorithm>#include<iomanip>#include<vector>#include<string>#include<bitset>#include<queue>#include<map>#include<set>using namespace std;inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*f;}void print(int x){if(x<0)putchar('-'),x=-x;if(x>=10)print(x/10);putchar(x%10+'0');}const int N=50010;int con[N],sum[N],last[7];int main(){int n=read();register int i,x,ans=0;for(i=1;i<7;++i)last[i]=n;for(i=1;i<=n;++i){con[i]=con[i-1];sum[i]=sum[i-1];x=read();if(x){x%=7;con[i]++;(sum[i]+=x)%=7;}last[sum[i]]=min(last[sum[i]],i);if(con[i]-con[last[sum[i]]]>0)ans=max(ans,i-last[sum[i]]);}print(ans);puts("");return 0;}