Description
This PRoblem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.
An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,… are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.
As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.
For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.
Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it’s the product of three H-primes.
Input
Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.
Output
For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.
Sample Input
21 85 789 0 Sample Output
21 0 85 5 789 62
大致题意: 一个H-number是所有的模四余一的数。 如果一个H-number是H-primes 当且仅当它的因数只有1和它本身(除1外)。 一个H-number是H-semi-prime当且仅当它只由两个H-primes的乘积表示。 H-number剩下其他的数均为H-composite。 给你一个数h,问1到h有多少个H-semi-prime数。
直接打表。。。 有一个问题,我到现在才发现if(-1)的判断也为真。。。非0即为真,写了两年C++连这个都不清楚,真鸡儿丢人
#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>//#include <map>#include <cmath>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 1000005#define Mod 10001using namespace std;int vis[MAXN],ans[MAXN];void Init(){ memset(vis,0,sizeof(vis)); for(int i=5; i<=MAXN; i+=4) for(int j=5; j<=MAXN; j+=4) { int tmp=i*j; if(tmp>MAXN) break; if(vis[i]==0&&vis[j]==0) vis[tmp]=1; else vis[tmp]=-1; } int cnt=0; for(int i=1; i<=MAXN; ++i) { if(vis[i]==1) cnt++; ans[i]=cnt; }}int main(){ Init(); int h; while(~scanf("%d",&h)&&h) { printf("%d %d/n",h,ans[h]); } return 0;}新闻热点
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