B. Hungry Sequence
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the waiter wants Iahub to write a Hungry sequence consisting of n integers.
A sequence a1, a2, ..., an, consisting of n integers, is Hungry if and only if:
- Its elements are in increasing order. That is an inequality ai < aj holds for any two indices i, j (i < j).
- For any two indices i and j (i < j), aj must not be divisible by ai.
Iahub is in trouble, so he asks you for help. Find a Hungry sequence with n elements.
Input
The input contains a single integer: n (1 ≤ n ≤ 105).
Output
Output a line that contains n space-separated integers a1 a2, ..., an (1 ≤ ai ≤ 107), representing a possible Hungry sequence. Note, that each ai must not be greater than 10000000 (107) and less than 1.
If there are multiple solutions you can output any one.
Sample test(s)
input
3
output
2 9 15
input
5
output
11 14 20 27 31
//我火星了，其实从100000,开始输出就可以了，但我。。。唉
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define N 1000000
int a[N+1]={0},p[N]={2,3};
int main(){
int n,m=2,i,j,k=4;
int x,y;
for(i=5;i<N;i+=k^=6){
if(a[i]==0){
p[m++]=i;
for(j=i;j<=N/i;j++)a[j*i]=-1;
}
}
//for(i=0;i<100;i++)cout<<p[i]<<" ";
//cout<<m<<endl;
while(~scanf("%d",&n)){
if(n<=m){
printf("%d",p[0]);
for(i=1;i<n;i++)printf(" %d",p[i]);printf("\n");
}else{
printf("10");//5,7,11,....
for(i=j=4,x=14,y=15;--n;){
if(x>y){
printf(" %d",y);
y=p[j++]*3;
}else{
printf(" %d",x);
x=p[i++]*2;
}
}
printf("\n");
}
}
return 0;
}