D - Tokitsukaze and Multiple

https://vjudge.net/contest/386223#problem/D

Tokitsukaze has a sequence of length

n

n

InputThere are several test cases.

The first line contains an integer $T$ OutputFor each test case, output in one line the maximum possible number of elements that are multiples of $p$ Sample Input

2
5 3
2 1 3 2 1
3 1
123 456 789

Sample Output

3
3

Sponsor

题意：

　　一个序列，每次可以将序列相邻的两个数相加生成一个新的和替代这两个元素，求最终序列里为

思路：

　　前缀和

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=2e5+10;
const int mod=1e9+7;

int n = 0;
int p = 0;
int counter[N] = {0};
int sum[N] = {0};
int ans[N] = {0};
int K[N] = {0};
void init_K()
{
    for(int i = 0;i<p;i++)
	{
	 	K[i] = -1;
	}
}
int main()
{
    int T;
    cin >> T;
    while(T--)
	{
        scanf("%d%d",&n ,&p);
        init_K();
        for(int i = 1;i<=n;i++)
		{
			int temp = 0;
			cin >> temp;
        	counter[i] = (counter[i-1] + temp)%p;
        }
        for(int i = 0;i<=n;i++)
		{
            sum[i] = K[counter[i]];
            K[counter[i]] = i;
        }
        for(int i = 1;i<=n;i++)
		{
            if(sum[i] != -1)
            {
                ans[i] = max(ans[i-1] , ans[sum[i]] + 1);
            }
            else 
			{
			 	ans[i] = ans[i-1];
			}
        }
        printf("%d\n",ans[n]);
    }
    return 0;
}

附第一遍错误思路OT代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=2e5+10;
const int mod=1e9+7;

int counter[100005] = {0};
void init()
{
    for(int i = 0;i < 100005;i++)
		counter[i] = 0;
}
int main()
{
	int T = 0;
	int n  = 0, p = 0;
	int temp = 0;
	int ans = 0;
	cin >> T;
	while(T--)
	{
		init();
		ans = 0;
		cin >> n >> p;
		for(int i = 0;i<n;i++)
		{
			cin >> temp;
			counter[temp % p] += 1;
		}
		ans += counter[0];
		for(int i = 1;i <= p/2;i++)
		{
			ans += min(counter[i],counter[p-i]);
		}
		if(p % 2 == 0)
		{
			ans += counter[p/2]/2;
		}
		cout << ans << endl;
	} 
	return 0;
}

原文地址：https://www.cnblogs.com/SutsuharaYuki/p/13416909.html