Subsequences Summing to Sevens-白红宇

Subsequences Summing to Sevens

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发布时间：2019-06-07

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Subsequences Summing to Sevens

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.

输入

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

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.

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++.

样例输入

7351621410

样例输出

提示

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

分析：(i+j)%k=i%k,则j是k的倍数；

代码：

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               #include 
                
                 #include 
                 
                  #include 
                  
                   #include 
                   
                    #include 
                    
                     #define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set
                     
                      ::iterator it=s.begin();it!=s.end();it++)#define vi vector
                      
                       #define pii pair
                       
                        #define mod 1000000007#define inf 0x3f3f3f3f#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)const int maxn=1e6+10;const int dis[4][2]={ { 0,1},{-1,0},{ 0,-1},{ 1,0}};using namespace std;ll gcd(ll p,ll q){ return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){ if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int n,m;ll a[maxn],pre[7],last[7],now;int main(){ int i,j,k,t; scanf("%d",&n); rep(i,1,n){ scanf("%lld",&a[i]); now=(now+a[i])%7; if(pre[now])last[now]=i; else pre[now]=i; } ll ma=0; rep(i,0,6) { if(pre[i]&&last[i])ma=max(ma,last[i]-pre[i]); } printf("%lld\n",ma); //system ("pause"); return 0;}

转载于:https://www.cnblogs.com/dyzll/p/5743093.html

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