123 A. Prime Permutation
You are given a string s, consisting of small Latin letters. Let's denote the length of the string as |s|. The characters in the string are numbered starting from 1.
Your task is to find out if it is possible to rearrange characters in string s so that for any prime number p ≤ |s| and for any integer i ranging from 1 to |s| / p (inclusive) the following condition was fulfilled sp = sp × i. If the answer is positive, find one way to rearrange the characters.
The only line contains the initial string s, consisting of small Latin letters (1 ≤ |s| ≤ 1000).
If it is possible to rearrange the characters in the string so that the above-mentioned conditions were fulfilled, then print in the first line "YES" (without the quotes) and print on the second line one of the possible resulting strings. If such permutation is impossible to perform, then print the single string "NO".
abc
YES
abc
abcd
NO
xxxyxxx
YES
xxxxxxy
In the first sample any of the six possible strings will do: "abc", "acb", "bac", "bca", "cab" or "cba".
In the second sample no letter permutation will satisfy the condition at p = 2 (s2 = s4).
In the third test any string where character "y" doesn't occupy positions 2, 3, 4, 6 will be valid.
题目大意
输入一个字符串，判断能否使每一质数和其倍数位上的字符均相同，能输出YES和任一合法解，否则输出NO
分析
一、例如6即使3的倍数又是2的倍数，所以s[3]=s[6],s[2]=s[6],所以这三位要相同。那我们不妨用并查集来维护一组需要全部相同的下标，父亲记作最小的点。要实现这个目的，我们要先处理出全部质数和其倍数，然后判断1到n的所有质数中如果有一质数i乘另一质数j小于n就将j连在i上。经过这一步可以判断出所有要求字符相同的下标位置。
二、下一步就是判断是否合法。我们不妨列几个数（1、2、3、4、5、6、7、8、9、10），我们不难发现不存在次大集合下标数加第三大集合下标数大于最大集合下标数的可能性，所以我们可以放心大胆的将数量最多的那个字母给最大集合中的这些下表用，然后减掉这些数后重新给各个字母的数量排序（字母一共只有26个，所以不用担心超时）。在判断是否合法时顺便将给各集合的字母记录一下，如果中间走不下去了输出NO，否则输出YES然后把之前记录下的字母按下标输出就大功告成了。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int prime[5000],fa[5000],n,t[500000];
char pr[500000];
struct node{
int t,pl;
}num[50];
struct hhh{
int t,pl;
}sum[100000];
void init()
{ int i,j,k;
memset(prime,1,sizeof(prime));
prime[1]=prime[0]=0;
fa[1]=1;
fa[0]=0;
for(i=2;i<=1100;i++)
if(prime[i]){
fa[i]=i;
for(j=i*2;j<=1100;j+=i){
prime[j]=0;
if(!fa[j]);
fa[j]=i;
}
}
}
int sf(int x)
{ return (fa[x]==x?x:fa[x]=sf(fa[x]));
}
void merge()
{ int i,j,k;
for(i=2;i<=n;i++)
if(prime[i])
for(j=i+1;j<=n;j++)
if(prime[j])
if(i*j<=n&&fa[i]!=fa[j]){
int x,y;
x=sf(i);
y=sf(j);
if(x!=y)
fa[y]=x;
}
}
bool cmp(const node &x,const node &y)
{ return x.t>y.t;
}
bool cmp2(const hhh &x,const hhh &y)
{ return x.t>y.t;
}
int main()
{ int m,i,j,k;
string s;
cin>>s;
n=s.size();
for(i=0;i<n;i++)
num[s[i]-'a'+1].t++;
init();
merge();
for(i=1;i<=27;i++)
num[i].pl=i;
for(i=1;i<=n;i++)
sum[sf(i)].t++;
for(i=1;i<=n;i++)
sum[i].pl=i;
sort(sum+1,sum+n+1,cmp2);
sort(num+1,num+27,cmp);
k=1;
while(num[1].t){
if(!sum[k].t)continue;
if(num[1].t<sum[k].t){
puts("NO");
return 0;
}
num[1].t-=sum[k].t;
t[sum[k].pl]=num[1].pl;
k++;
sort(num+1,num+27,cmp);
}
puts("YES");
for(i=1;i<=n;i++)
pr[i]=t[fa[i]]-1+'a';
for(i=1;i<=n;i++)
cout<<pr[i];
cout<<endl;
return 0;
}
