标签：printf lag output with crypto spoj ret 个数 peter
求给定的两个数之间的素数
Translated by @kaiming
Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers!
输入格式：
The input begins with the number t of test cases in a single line (t<=10). In each of the next t lines there are two numbers m and n (1 <= m <= n <= 1000000000, n-m<=100000) separated by a space.
输出格式：
For every test case print all prime numbers p such that m <= p <= n, one number per line, test cases separated by an empty line.
After cluster change, please consider PRINT as a more challenging problem.
//Pro: Spoj PRIME1 - Prime Generator //求给定的两个数之间的素数 #include<iostream> #include<cmath> #include<cstdio> #include<algorithm> using namespace std; const long long N=40000; long long T; long long cnt; long long prime[N+1]; bool flag[N+1]; void init() { flag[1]=1; long long tmp; for(long long i=2;i<=N;++i) { if(!flag[i]) prime[++cnt]=i; for(long long j=1;j<=cnt&&(tmp=i*prime[j])<=N;++j) { flag[tmp]=1; if(i%prime[j]==0) break; } } } long long l,r; bool Flag; int main() { init(); scanf("%d",&T); while(T--) { scanf("%d%d",&l,&r); for(;l<=r;++l) { if(l<=N) { if(!flag[l]) { printf("%d\n",l); continue; } } else { Flag=0; for(long long j=1;prime[j]<=sqrt(l)&& !Flag && j<=cnt;++j) { if(l%prime[j]==0) Flag=1; } if(!Flag) printf("%d\n",l); } } puts(""); } return 0; }
标签：printf lag output with crypto spoj ret 个数 peter
原文地址：https://www.cnblogs.com/lovewhy/p/9246205.html
