poj 1595

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ACM/ICPC 之数论-素数筛选法与 "打表"思路(POJ 1595)

何为"打表"呢,说得简单点就是: 有时候与其重复运行同样的算法得出答案,还不如直接用算法把这组数据所有可能的答案都枚举出来存到一个足够大的容器中去-例如数组(打表),然后再输入数据的时候,直接遍历容器,检索这个数据是否有题意要求的结果. 举一个几乎所有程序员都知道的简单例子= =: 求素数(POJ 1595)-Prime cuts 这一题大意是给出多组N(1~1000)和C,让你从N内素数的中间项向外扩展C个素数,比如给出7 1,素数有5个(注意此题出题人坑爹得让1作为"素…

#include <iostream> #define N 10010 using namespace std; int a[N],b[N]; int prime(int a) { int i; for(i=2;i*i<=a;i++) if(a%i==0) return 1; return 0; } int main() { int i,j,t,m,x,h,g; while(cin>>t>>x){ h=t;g=x; int k=1; for(m=t;m>1;…

POJ 1595 Prime Cuts （ZOJ 1312）素数打表

ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=312 POJ:http://poj.org/problem?id=1595 题目大意: 给你两个数n和c,如果1~n(包括1和n)之间的素数个数为偶数个,则输出中间c*2个素数否则输出中间c*2-1个素数. 思路: 呀呀呀,玩了一天了,A道水题洗洗睡吧. 直接筛选素数打表. 注意c比1~n的个数大的情况要把1~n中所以素数输出. OK,然后这题1也算素数.. #incl…

POJ 1595 素数打表水题

[题意简述]:给出N和C,让我们求出N以内的包含N的素数,然后依据若N以内的素数为奇数个,就将中间2*c-1个素数输出:若为偶数个.就将中间2*c个素数输出. [分析]:仅仅要题意理解就简单了. 详见代码: // 224K 16Ms #include<iostream> using namespace std; #define N 2000 bool isprime[N]; int prime[N],nprime; void doprime(int n) { int i,j; nprime =…

PoJ 1595 PrimeCuts

Prime Cuts Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9339 Accepted: 3562 Description A prime number is a counting number (1, 2, 3, ...) that is evenly divisible only by 1 and itself. In this problem you are to write a program tha…

poj 1595 Prime Cuts

Prime Cuts Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 10610 Accepted: 4046 Description A prime number is a counting number (1, 2, 3, ...) that is evenly divisible only by 1 and itself. In this problem you are to write a program th…

[暑假集训--数论]poj1595 Prime Cuts

A prime number is a counting number (1, 2, 3, ...) that is evenly divisible only by 1 and itself. In this problem you are to write a program that will cut some number of prime numbers from the list of prime numbers between (and including) 1 and N. Yo…