Necklace CodeForces - 613C (构造)

链接

大意: 给定n种珠子, 第i种有$a_i$个, 求将珠子穿成项链, 使得能使切开后的项链回文的切口尽量多

若有一种以上珠子为奇数, 显然不能为回文, 否则最大值一定是$gcd(a_1,a_2,...,a_n)$

若有一个奇数, 直接分成$gcd(a_1,a_2,...,a_n)$块, 每块内奇数放中间, 其余对称分

无奇数的话, $gcd(a_1,a_2,...,a_n)$一定是2的倍数, 可以将2块和为1块, 两块间对称分即可

#include <iostream>
#include <algorithm>
#include <cstdio>
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

const int N = 4e5+10, INF = 0x3f3f3f3f;
int n, cnt, ans, sum;
int a[N];
char buf[N];

int main() {
	scanf("%d", &n);
	REP(i,1,n) {
		scanf("%d", a+i);
		ans = gcd(ans, a[i]);
		cnt += a[i]&1;
		sum += a[i];
	}
	if (cnt>1) {
		puts("0");
		REP(i,1,n) REP(j,1,a[i]) putchar(i-1+'a');
		return puts(""),0;
	}
	if (cnt) {
		int len = sum/ans, cur = len/2+1;
		REP(i,1,n) if (a[i]&1) {
			buf[cur] = i-1+'a';
			++cur; break;
		}
		REP(i,1,n) REP(j,1,a[i]/ans/2) buf[cur]=buf[len-cur+1]=i-1+'a',++cur;
		printf("%d\n", ans);
		REP(i,1,ans) REP(j,1,len) putchar(buf[j]);
		return puts(""),0;
	}
	int len = sum/ans*2, cur = len/2+1;
	REP(i,1,n) REP(j,1,a[i]/ans) buf[cur]=buf[len-cur+1]=i-1+'a',++cur;
	printf("%d\n", ans);
	REP(i,1,ans/2) REP(j,1,len) putchar(buf[j]);
	puts("");
}