CF 613C(Necklace-构造法)

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给你n (1 ≤ n ≤ 26)种颜色的珠子，每种ai个，你需要把它们全部使用连成环，使其从‘某处’断开后形成的串是回文的，这样的某处尽可能多。求出任意一种方案

构造

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <map>#include <functional>#include <cstdlib>#include <queue>#include <stack>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define MEM(s) memset(s,0,sizeof(s));#define Fork(i,k,n) for(int i=k;i<=n;i++) #define Forp(x) for(int p=pre[x];p;p=next[p])  #define RepD(i,n) for(int i=n;i>=0;i--)  #define pb push_backconst int maxn = 100000 + 10;int n,a[maxn];vector<int> v1,v2;int gcd(int a,int b){if (!b) return a;return gcd(b,a%b);}void pri(){    For(i,n)         For(j,a[i]) printf("%c",i+'a'-1);    cout<<endl;}int main() {    cin>>n;    int tot=0;    For(i,n) {        cin>>a[i];        tot+=a[i];        if (a[i]&1) v1.pb(i);        else if (a[i]) v2.pb(i);    }    int m1=v1.size(),m2=v2.size();    if (m1+m2==1)    {        cout<<tot<<endl;        pri();        return 0;    }    int g=0;    For(i,n) g=gcd(a[i],g);    if (tot&1) {        if (m1>=2) {            cout<<"0"<<endl;            pri();            return 0;                   }        cout<<g<<endl;        a[v1[0]]-=g;        For(k,g)        {                For(i,n) For(j,a[i]/g/2) printf("%c",i+'a'-1);                printf("%c",v1[0]+'a'-1);                       ForD(i,n) For(j,a[i]/g/2) printf("%c",i+'a'-1);        }        cout<<endl;    } else {        if (m1) {            cout<<"0"<<endl;            pri();            return 0;                   }        cout<<g<<endl;        For(k,g)        {                if (k&1) For(i,n) For(j,a[i]/g) printf("%c",i+'a'-1);                else ForD(i,n) For(j,a[i]/g) printf("%c",i+'a'-1);              }        cout<<endl;    }    return 0;}

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