题目链接

题面:

题意:
给定一个长度为 n 的数列 a,a 数列中的相邻的两个数可以合并,问最多可以让数列 a 中有多少个数是p的倍数。

题解:
我们把 a 数列中某一段相加是p的倍数的段都拿出来,这样问题转化为,给定一些线段,选出最多的不相交的线段。

代码:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<bitset>
#include<map>
#include<unordered_map>
#include<set>
#include<list>
#define ui unsigned int
#define ll long long
#define llu unsigned ll
#define ld long double
#define pr make_pair
#define pb push_back
#define lc (cnt<<1)
#define rc (cnt<<1|1)
//#define len(x) (t[(x)].r-t[(x)].l+1)
#define tmid ((l+r)>>1)
#define max(x,y) (x)>(y)?(x):(y)
#define min(x,y) (x)>(y)?(y):(x)
using namespace std;

const int inf=0x3f3f3f3f;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const double dnf=1e18;
const int mod=1e9+7;
const double eps=1e-1;
const double pi=acos(-1.0);
const int hp=13331;
const int maxn=100100;
const int maxp=1100;
const int maxm=4000100;
const int up=1000;

int a[maxn],dp[maxn],pos[maxn];

int main(void)
{
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        int n,p;
        scanf("%d%d",&n,&p);
        for(int i=0;i<=p;i++)
            pos[i]=0;

        int maxx=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]=(a[i-1]+a[i])%p;
            if(pos[a[i]]||a[i]==0)
                dp[i]=max(dp[pos[a[i]]]+1,dp[i-1]);
            else
                dp[i]=max(0,dp[i-1]);
            pos[a[i]]=i;
            maxx=max(maxx,dp[i]);
        }
        printf("%d\n",maxx);
    }
    return 0;
}