[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[];

long long p[];

long long x,l,ans;

void in(long long &x)

{

    char c=g();x=;

    while(c<''||c>'')c=g();

    while(c<=''&&c>='')x=x*+c-'',c=g();

}

void o(long long x)

{

    if(x>)o(x/);

    p(x%+'');

}

int main()

{

    in(n);

    For(i,,n)

    {

        in(x);

        a[i]=x+a[i-];

    }

/*    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,,)

    p[i]=inf;

    For(i,,n)

    {

        p[a[i]%]=min(p[a[i]%],i);

        ans=max(ans,i-p[a[i]%]);

    }

    o(ans);

     return ;

}