2020杭电多校第三场 Tokitsukaze and Multiple

第四题：Tokitsukaze and Rescue

题目：

Tokitsukaze has a sequence of length n, denoted by a.

Tokitsukaze can merge two consecutive elements of a as many times as she wants. After each operation, a new element that equals to the sum of the two old elements will replace them, and thus the length of a will be reduced by 1.

Tokitsukaze wants to know the maximum possible number of elements that are multiples of p she can get after doing some operations (or doing nothing) on the sequence a.

思路

将前缀和的模数存进栈，如果栈内有两个数相等，那么就意味着这之间的数的和可以为p的倍数，那么我们就把这写数合并，因为合并了 前面的数也不能用了，所以就把栈清空，假设我们那个数模p为x，那么如果后面有个人膜了p之后等于x就证明这之间的也可以用。 所以我们要把x进栈。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;

const int N=1e5+10;
ll a[N],sum[N]; 
int stk[N],top;
int main()
{
    //freopen("test.in","r",stdin);//设置 cin scanf 这些输入流都从 test.in中读取
    //freopen("test.out","w",stdout);//设置 cout printf 这些输出流都输出到 test.out里面去
    //ios::sync_with_stdio(false);
    //cin.tie(0),cout.tie(0);
    int T;
    cin>>T;
    while(T--)
    {
        memset(stk,0,sizeof stk);
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]),sum[i]=sum[i-1]+a[i]; 
        top=0;
        int ans=0;
        stk[top++]=0;
        for(int i=1;i<=n;i++)
        {
            stk[top++]=sum[i]%m;
            for(int j=0;j<top-1;j++)
            {
                if(stk[j]==stk[top-1])
                {
                    ans++;
                    stk[0]=stk[top-1];
					top=1;
                    break;
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}