[USACO16JAN]子共七Subsequences Summing to Sevens

[USACO16JAN]子共七Subsequences Summing to Sevens
a[i]表示前缀和
如果a[i]%7==t&&a[j]%7==t
那么a[j]-a[i-1]一定是7的整数倍，
这样就o(n)扫一遍，不断更新答案就可以了。

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<cstring>
#define inf 2147483647
#define For(i,a,b) for(register long long i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()
//by war
//2017.10.12
using namespace std;
long long n;
long long a[50010];
long long p[50010];
long long x,l,ans;
void in(long long &x)
{
    char c=g();x=0;
    while(c<'0'||c>'9')c=g();
    while(c<='9'&&c>='0')x=x*10+c-'0',c=g();
}
void o(long long x)
{
    if(x>9)o(x/10);
    p(x%10+'0');
}
int main()
{
    in(n);
    For(i,1,n)
    {
        in(x);
        a[i]=x+a[i-1];
    }
/*    for(register long long len=n;len>=1;len--)
    {
        For(i,1,n-len+1)
        if((a[i+len-1]-a[i-1])%7==0)
        {
        o(len);    
        exit(0);
        }
    }*/
    For(i,0,6)
    p[i]=inf;
    For(i,1,n)
    {
        p[a[i]%7]=min(p[a[i]%7],i);
        ans=max(ans,i-p[a[i]%7]);
    }
    o(ans);
     return 0;
}