Misha and Palindrome Degree CodeForces - 501E (回文串计数)

大意: 给定字符串, 求多少个区间重排后能使原串为回文串.

先特判掉特殊情况, 对于两侧已经相等的位置之间可以任意组合, 并且区间两端点至少有一个在两侧相等的位置处, 对左右两种情况分别求出即可.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m, k, t;
int f[N], a[N], c[N];

int main() {
	a[0] = -1;
	scanf("%d", &n);
	REP(i,1,n) scanf("%d",a+i),++f[a[i]];
	REP(i,1,n) m += f[i]&1;
	if (m>1) return puts("0"),0;
	int pos = 1, len = 0;
	while (a[pos]==a[n-pos+1]) f[a[pos++]]-=2;
	if (pos>n) return printf("%lld\n",(ll)n*(n+1)/2),0;
	PER(i,1,n-pos+1) {
		if (2*++c[a[i]]>f[a[i]]) {
			if (i>n-i+1||a[i]!=a[n-i+1]) break;
			if (i==n-i+1&&f[a[i]]%2==0) break;
		}
		++len;
	}
	memset(c,0,sizeof c);
	REP(i,pos,n) {
		if (2*++c[a[i]]>f[a[i]]) {
			if (i<n-i+1||a[i]!=a[n-i+1]) break;
			if (i==n-i+1&&f[a[i]]%2==0) break;
		}
		++len;
	}
	printf("%lld\n", (ll)pos*(pos+len));
}