标签：get ted led 题意 long spec c++ 打印 hold
题目链接：http://codeforces.com/contest/1174/problem/C
You‘re given an integer n. For every integer i from 2 to n, assign a positive integer ai such that the following conditions hold:
For any pair of integers (i,j), if i and j are coprime, ai≠aj
The maximal value of all ai should be minimized (that is, as small as possible).
A pair of integers is called coprime if their greatest common divisor is 1.
The only line contains the integer n (2≤n≤105).
Print n−1
integers, a2, a3, …, an (1≤ai≤n).
If there are multiple solutions, print any of them.
给你一个数，让你输出长度为n-1的数组，这个数组的起始下标从2开始，使每组任意下标互质的两个数所对应的值都互质。
由于范围在10^5，故可以打素数表，遇到2的倍数打印1，遇到3的倍数打印2，遇到5的倍数打印3...遇到质数的倍数打印该质数在质数表中的位置即可。
// // Created by hjy on 19-6-4. // #include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=2e5+10; int n; bool prime(int m)//简单判断素数 { for(int i=2;i<=sqrt(m);i++) { if(m%i==0) return false; } return true; } int result[maxn]={0}; void cun()///存入表中 { int op=0; for(int i=2;i<=maxn;i++) { if(prime(i)) { op++; //cout<<"i="<<i<<endl; for(int j=i;j<=maxn;j+=i) { //cout<<"j="<<j<<endl; result[j]=op; //cout<<"result[j]="<<result[j]<<endl; } } } } int main() { int n; while(cin>>n) { cun(); for(int i=2;i<=n;i++) cout<<result[i]<<‘ ‘; cout<<endl; } return 0; }
Codeforces Round #563 (Div. 2)C
标签：get ted led 题意 long spec c++ 打印 hold
原文地址：https://www.cnblogs.com/Vampire6/p/10990359.html
