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SPOJ PRIME1 - Prime Generator(线性筛)

Peter wants to generate some prime numbers for his cryptosystem. Help him!...Your task is to generate all prime numbers between two given numbers!...vis[i]) prime[++tot] = i; for(int j = 1; j prime[j] * i prime[j]] = 1; if(!...0; for(int i = 1; prime[i] <= limit; i++) if((x % prime[i]) == 0) return 0; return 1

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poj 3126 Prime Path

I am the Prime minister, you know! — I know, so therefore your new number 8179 is also a prime....Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime! ...— I see, being the prime minister you cannot stand having a non-prime number on your door even for a...is changed from one prime to the next prime. ...Help the prime minister to find the cheapest prime path between any two given four-digit primes!

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POJ 1365 Prime Land

: 979 Description Everybody in the Prime Land is using a prime base number system....In fact, the children in Prime Land learn to add to subtract numbers several years....Help people in the Prime Land and write a corresponding program. ...from the prime base representation above for which ei > 0....while(ans%prime[i]==0) { cas++; ans/=prime[i];

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ACMSGURU 231 - Prime Sum

Prime Sum Problem Description Find all pairs of prime numbers (A, B) such that Aprime number and does not exceed N....(); i++) { if(prime[i] - prime[i - 1] == 2) { count += 1; } } std...::cout << count << std::endl; for(unsigned long i = 1; i prime.size(); i++) { if(prime...[i] - prime[i - 1] == 2) { std::cout prime[i - 1] << std::endl; }

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牛式 Prime Cryptarithm

P1211 [USACO1.3]牛式 Prime Cryptarithm 分析：大暴力，还说啥呢...标记输入的数字，一个个算出来，判断是否有标记到 #include using

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【USACO 1.5】Prime Palindromes

10 */ 11 #include 12 #include 13 int l,r,ans; 14 bool is_prime(int x){ 15 for(int...if(d==(len+1)/2){ 21 int x; 22 sscanf(s,"%d",&x); 23 if(x>=l&&xprime

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poj Anti-prime Sequences

Anti-prime Sequences Time Limit: 3000MS Memory Limit: 30000K Total Submissions: 2175 Accepted: 1022...sums to a composite (non-prime) number....For example, if n = 1 and m = 10, one such anti-prime sequence is 1,3,5,4,2,6,9,7,8,10....In the case where no anti-prime sequence exists, output No anti-prime sequence exists. ...sequence exists. 40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54 题意：求n到m的数中任意连续2到d的数的和是合数

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Prime Ring Problem hdu 1016

., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime....以下是AC代码 #include #include #include using namespace std; int prime[50...],book[50]; int ans[50],n; void pri()//筛选素数还有更快的改进版本，有兴趣可以百度 { prime[0]=prime[1]=0; prime[2]=1; for...(int i=2;i<=50;i++) { for(int j=i*i;j<=50;j+=i) prime[j]=0; } } void dfs(int in)...[i]=0; } } } int main() { int cnt=1; fill(prime,prime+50,1); pri(); while(cin>>

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Prime Path （POJ - 3126 ）（BFS）

分别要从个位、十位、百位、千位判断，在个位的时候每次只能是1、3、5、7、9，其他的改变之后都不是素数，十位、百位、千位都从0开始遍历到9，每次只要符合是素数就放到队列中，开个结构体记录步数和当前的数就可以了...I am the Prime minister, you know! — I know, so therefore your new number 8179 is also a prime....— I see, being the prime minister you cannot stand having a non-prime number on your door even for a...is changed from one prime to the next prime....Help the prime minister to find the cheapest prime path between any two given four-digit primes!

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ACMSGURU 113 - Nearly prime numbers

Nearly prime numbers Problem Description Nearly prime number is an integer positive number for which...N integer positive numbers, you are to write a program that prints “Yes” if given number is nearly prime...Write “Yes” in it if given number is nearly prime and “No” in other case....(), prime.end(), num / primeNum); if(res !...= prime.end() && *res * primeNum == num) { flag = true; break

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POJ 3126 Prime Path(bfs)

std; const int MAXN = 10000; struct Node{ int num,step; }Now,Next; int vis[MAXN]; int s,e,n; bool Prime...[MAXN]; void prime(){ // 将素数记录下来 int m = sqrt(MAXN); memset(Prime,true,sizeof(Prime));...Prime[0] = Prime[1] = false; for(int i=2;i<m;i++){ if(Prime[i]){ for(int j=i*i;jPrime[j] = false; } } } } bool judge(int a,int b){ //判断是否只改变了一个数字 int ans = 0; if(...("%d\n",temp); } else { printf("Impossible\n"); } } return 0; } /* [来源] POJ 3126 [题目] Prime

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Prime Number of Set Bits in Binary Representation

Also, 1 is not a prime.)...7 -> 111 (3 set bits, 3 is prime) 9 -> 1001 (2 set bits , 2 is prime) 10->1010 (2 set bits...bits, 2 is prime) 11 -> 1011 (3 set bits, 3 is prime) 12 -> 1100 (2 set bits, 2 is prime)...思路：判断二进制中1的个数是否为素数，int型总共32位，所以只需要枚举出中的数枚举即可。...Java版本： public int countPrimeSetBits(int L, int R) { boolean[] isPrime = new boolean[32];

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回文质数 Prime Palindromes

101 131 151 181 191 313 353 373 383 说明 Hint 1: Generate the palindromes and see if they are prime...using namespace std; const int maxn=9989999; //小小的cheat 打表看到最大是9989899 bool isprime[maxn]; void prime...(int o); bool hw(string tem); int main() { int a,b; cin>>a>>b; if(b>maxn) b=maxn-1; prime(b);...bool hw(string tem) { string w=tem; reverse(w.begin(),w.end()); return (w==tem); } void prime

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poj 3126 Prime Path （广搜）

[MAXN]; int vis[MAXN]; int step[MAXN]; void init() { memset(vis_prime,0,sizeof(vis_prime));...for(int i=2; i<=(int)sqrt(1.0*MAXN); i++) { if(vis_prime[i]==0) { for...(int j=i*2; j<MAXN; j+=i) { vis_prime[j]=1; } }...} //for(int i=1000;iprime[i]==0) printf("%d ",i); } int BFS(int a,int b) {...vis_prime[next] && !

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07:清泉-改（prime+堆）

) 40 {p=a;v=b;} 41 bool operator<(const pr&a)const 42 {return v>a.v;} 43 }inc; 44 void prime...read(x);read(y);read(z); 82 add_edge(x,y,z); 83 add_edge(y,x,z); 84 } 85 prime

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K-th Smallest Prime Fraction

K-th Smallest Prime Fraction Problem: A sorted list A contains 1, plus some number of primes.

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数学--数论--POJ1365——Prime Land

Description Everybody in the Prime Land is using a prime base number system....,e1, e0) is considered to be the representation of x in prime base number system....In fact, the children in Prime Land learn to add to subtract numbers several years....Help people in the Prime Land and write a corresponding program....from the prime base representation above for which ei > 0.

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获取N以内的质数(Prime)

方法一： /** * @Author Diuut * @Date 2019/8/15 20:54 */ public class Test0814_1_Prime { public static...System.currentTimeMillis(); System.out.println("本次运行耗时:"+(end-start)); } } 本次运行耗时:3736 方法二 import java.util.ArrayList...; import java.util.List; //检验一个数是不是素数，如果所有小于它的素数都不能将它整除，那么它就是素数 public class Test0814_2_Prime { public...方法三 import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; /** * @Author...用户访问的时候直接读取缓存好的静态页面，而不是发送请求，服务器再从数据库中读取，简单暴力没有技术含量，但却有效。 Post Views: 81

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HDUOJ----(1016)Prime Ring Problem

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others..., 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime

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POJ 3581 Prime Gap（二分）

Prime Gap Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 11009 Accepted: 6298 Description...to 1) lying between two successive prime numbers p and p + n is called a prime gap of length n....For example, ‹24, 25, 26, 27, 28› between 23 and 29 is a prime gap of length 6....For convenience, the length is considered 0 in case no prime gap contains k....} else lo=tem; if(lo+1==up)break; } printf("%d\n",prime[up]-prime

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