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.
InputThe input contains a single integer: n (1 ≤ n ≤ 105).
OutputOutput 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 题意：要求生成一个含n个数的数列，对于数列要求：1.升序。2.数列中的随意两个数互质。 这尼玛就是要输出素数嘛。。
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <cctype> #include <cstdlib> #include <set> #include <map> #include <vector> #include <string> #include <queue> #include <stack> #include <cmath> using namespace std; const int INF=0x3f3f3f3f; #define LL long long int prime[10000010],vis[10000010],num; void init() { memset(vis,1,sizeof(vis)); num=0; for(int i=2;i<10000000;i++) { if(vis[i]) { prime[num++]=i; for(int j=2;j*i<10000000;j++) vis[j*i]=0; } } } int main() { init(); int n; while(~scanf("%d",&n)) { for(int i=0;i<n;i++) if(i!=n-1) printf("%d ",prime[i]); else printf("%d\n",prime[i]); } return 0; }版权声明：本文博客原创文章。博客，未经同意，不得转载。
转载于:https://www.cnblogs.com/bhlsheji/p/4642809.html
相关资源：数据结构—成绩单生成器