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
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）
#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;
}
/*
*/
标签：Educational,Rated,141,300,Codeforces,int,solve,数组,Div From： https://www.cnblogs.com/cfddfc/p/17038695.html