Educational Codeforces Round 141 (Rated for Div. 2) Different Arrays

• Problem - D - Codeforces

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题意：

• 给一个长度为n的数组a，下标从2到n-1，每个位置执行一次操作
• 操作：设操作位置为i，$a_{i-1} += a_i, a_{i+1} -= a_i$，或者$a_{i-1} -= a_i, a_{i+1} += a_i$
• 问最终能得到多少个不同的数组a，数量模998244353
• 数据
  
  

   

   

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思路：

• 看到这个数据，有动态规划的感觉，但是对状态定义感觉很麻烦，所以就没做了
• 状态定义：$f[i][j]$表示数组a上第i个位置，元素大小为j的方案数
• 状态转移：
  • 当 j 不为0的时候
  • $f[i+1][j + a[i+1]] = \sum_{k=-M}^{M} f[i][j]$
  • $f[i+1][- j + a[i+1]] =\sum_{k=-M}^{M} f[i][j]$
  • 当 j 为0的时候，如果按照上面的方法，显然多算了一次（因为0不会改变数组）
  • $f[i+1][a[i+1]] = f[i][j]$
• 因为减法操作 j 的状态可能为负数，所以加上一个 M = 300*300，改变相对位置
• 数组大小：f[N][2*M],N=300,M=300*300
• 时间复杂度：O（N*M）

 

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#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<"  "<<#e<<" = "<<e<<endl;
#define endl "\n"
#define fi first
#define se second

//#define int long long
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;

//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)
const int N = 310,mod = 998244353,M = 300*300;
int a[N];
int f[N][M*2 + 10];

void solve() 
{
    int n;cin >> n;
    for(int i = 1;i <= n;i ++)cin >> a[i];

    f[2][a[2] + M] = 1;
    for(int i = 2;i <= n-1;i ++)
    {
        for(int j = -M;j <= M;j ++)
        {
            if(j)
            {
                if(j + a[i+1] + M >= 0 && j + a[i+1] + M <= 2*M)(f[i+1][j + a[i+1] + M] += f[i][j + M]) %= mod;
                if(-j + a[i+1] + M >= 0 && -j + a[i+1] + M <= 2*M)(f[i+1][-j + a[i+1] + M] += f[i][j + M]) %= mod;
            }else{
                f[i+1][a[i+1] + M] = f[i][j + M];
            }
        }
    }

    int ans = 0;
    for(int i = -M;i <= M;i ++)
    {
        (ans += f[n][i+M]) %= mod;
    }
    cout << ans << endl;
}

signed main()
{
    /*
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    */
    int T = 1;//cin >> T;

    while(T--){
        //puts(solve()?"YES":"NO");
        solve();
    }
    return 0;

}
/*

*/