reverse后差分循环同构：CF1045B

本文搬运自本人高中时期CSDN博客，若图片加载不出来，可到原文查看：https://blog.csdn.net/zhangtingxiqwq/article/details/135333770

https://www.luogu.com.cn/problem/CF1045B

分析题目可得，一个 xx 不能被凑出的充要条件是：

∀a∈A\forall a \in A ， ∃ a′∈A\exist\, a'\in A ，满足： a+a′≡b(modM)a+a'\equiv b\pmod M

显然 aa 变大时， a′a' 会变小。对 aa 排序，然后reverse为 a′a' 。如果 aa 和 a′a' 差分数组循环同构，则我们找出来一个 xx

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pre coding at 
st coding at 9:29
st bugging at 9:41
passing at 9:48
fn blogging at 9:57
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#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL
 #define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else
 #define debug(...) void(0)
#endif
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
#define fi first
#define se second
//#define M
#define mo 998244353
#define g (int)(1e9+7)
#define N 400010
int n, m, i, j, k, T;
map<int, int>mp; 
int s1[N], s2[N], c[N], a[N], b[N], M; 
int d1[N], d2[N]; 
vector<int>ans; 

void make(int *d, int n, int *s) {
	int i; 
	for(i = 1; i<n; ++i) s[i] = (s[i-1] * g + d[i] ) % mo; 
}

int qu(int l, int r, int *s) {
	return ( (s[r] - (s[l - 1] * c[r - l + 1])) % mo + mo ) %mo; 
}

signed main()
{
	#ifdef LOCAL
	  freopen("in.txt", "r", stdin);
	  freopen("out.txt", "w", stdout);
	#endif
//	srand(time(NULL));
//	T=read();
//	while(T--) {
//
//	}
	n=read(); M=read(); 
	for(i=1; i<=n; ++i) a[i]=read(); 
	sort(a+1, a+n+1);
	for(i=1; i<=n; ++i) b[i]=-a[n-i+1]; 
	for(i=1; i<=n; ++i) b[i+n]=b[i]+M; 
	for(i=1; i<=2*n; ++i) debug("%d ", b[i]); debug("\n"); 
	for(i=c[0]=1; i<=2*n; ++i) c[i]=c[i-1]*g%mo; 
	for(i=1; i<n; ++i) d1[i]=a[i+1]-a[i]; make(d1, n, s1); 
	for(i=1; i<2*n; ++i) d2[i]=b[i+1]-b[i]; make(d2, 2*n, s2); 
	for(i=1; i<n; ++i) debug("%d ", s1[i]); debug("\n"); 
	for(i=1; i<2*n; ++i) debug("%d ", s2[i]); debug("\n"); 
	for(i=1; i<=n+1; ++i) 
		if(qu(1, n-1, s1) == qu(i, i + n - 2, s2)) {
			k = ((a[1] - b[i]) % M + M) % M; 
			debug("%lld : %lld\n", i, k); 
			if(!mp[k]) ans.pb(k), mp[k]=1; 
		}
	printf("%d\n", ans.size()); 
	sort(ans.begin(), ans.end()); 
	for(auto t : ans) printf("%lld ", t); 
	return 0;
}