8 solutions
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2
#include<bits/stdc++.h> using namespace std; int x[10001]={0,1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039}; int main(){ int n; cin>>n; cout<<x[n+1]; return 0; }//致敬传奇打表王,可惜过不了,没读取道过程(没过程)。
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1
#include <bits/stdc++.h> using namespace std; int main(int argc, char **argv){ int n,zhi = 2; cin >> n; while (true){ bool flag = 1; for (int i = 2;i < zhi;i++){ if (i > 10000){ i *= i; } if (zhi % i == 0){ flag = 0; break; } } if (flag){ n--; } if (n == 0){ break; } zhi++; } cout << zhi; return 0; } //2 3 5 7 11 13 17 19 23 29
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0
#include<bits/stdc++.h> using namespace std; bool f; long long a; short n,sum=1; long long fn(int x){ for(int i=2;i<=INT_MAX;i++){ for(int k=2;k<i/2;k++){ if(i%k==0){ f=0; break;} } if(f==1){ a=i; sum++; f=0; if(sum==x) return a; }else f=1; } return 0; } int main(){ cin>>n; cout<<fn(n); return 0; }
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0
#include<bits/stdc++.h> using namespace std; bool sushu(int n){//判断n是否为质数 if(n<2) return false;//如果n<2 返回假 for(int i=2;i<=sqrt(n);i++){ if(n%i==0) return false;//如果n%i余数是0,说明n不是质数,返回假 } return true;//排除前两项,说明它是质数,返回真 } int main(){ int n,sum=0; cin>>n; for(int i=1;i>=1;i++){//循环遍历所有数 if(sushu(i)==true){//如果是质数 sum++;//就计数(贮存当前是第几个质数) } if(sum==n){//如果是第n小的质数,输出这个质数 cout<<i; return 0; } } return 0; }
- 1
Information
- ID
- 491
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 6
- Tags
- # Submissions
- 227
- Accepted
- 72
- Uploaded By