Codefoces 432 C. Prime Swaps

哥德巴赫猜想:

任一大于2的偶数，都可表示成两个素数之和。

任一大于5的整数都可写成三个质数之和。

贪心取尽可能大的素数.....

C. Prime Swaps time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

You have an array a[1], a[2], ..., a[n], containing distinct integers from 1 to n. Your task is to sort this array in increasing order with the following operation (you may need to apply it multiple times):

choose two indexes, i and j (1 ≤ i < j ≤ n; (j - i + 1) is a prime number); swap the elements on positions i and j; in other words, you are allowed to apply the following sequence of assignments: tmp = a[i], a[i] = a[j], a[j] = tmp (tmp is a temporary variable).

You do not need to minimize the number of used operations. However, you need to make sure that there are at most 5n operations.

Input

The first line contains integer n (1 ≤ n ≤ 105). The next line contains n distinct integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ n).

Output

In the first line, print integer k (0 ≤ k ≤ 5n) — the number of used operations. Next, print the operations. Each operation must be printed as "i j" (1 ≤ i < j ≤ n; (j - i + 1) is a prime).

If there are multiple answers, you can print any of them.

Sample test(s) input 3 3 2 1 output 1 1 3 input 2 1 2 output 0 input 4 4 2 3 1 output 3 2 4 1 2 2 4

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=100100; bool vis[maxn]; int prime[maxn/10],pn; int n,a[maxn],b[maxn]; int Left[maxn*5],Right[maxn*5],nu; void get_prime() { for(int i=2;i*i<maxn;i++) { for(int j=i*2;j<maxn;j+=i) { vis[j]=1; } } for(int i=2;i<maxn;i++) { if(vis[i]==0) prime[pn++]=i; } } int Bin(int x) { int low=0,high=pn-1,mid,ans=-1; while(low<=high) { mid=(low+high)/2; if(prime[mid]<=x) { ans=prime[mid],low=mid+1; } else high=mid-1; } return ans; } void debug() { cout<<"a....\n"; for(int i=1;i<=n;i++) cout<<a[i]<<","; cout<<endl; cout<<"b....\n"; for(int i=1;i<=n;i++) cout<<b[i]<<","; cout<<endl; } int main() { get_prime(); scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",a+i); b[a[i]]=i; } for(int i=1;i<=n;i++) { /// from b[i] to i greedy!!! int len=b[i]-i+1; while(len!=1) { int bin=Bin(len); int cp=b[i]-bin+1; /// changepos Left[nu]=cp,Right[nu]=b[i]; nu++; int t1=b[i],t2=b[a[cp]]; swap(a[cp],a[b[i]]); b[a[b[i]]]=t1;b[i]=t2; // debug(); getchar(); len=b[i]-i+1; } } printf("%d\n",nu); for(int i=0;i<nu;i++) { printf("%d %d\n",Left[i],Right[i]); } return 0; }

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