POJ 2773 happy2006 (素因子分解+容斥原理+二分)

来源：互联网发布：什么是软件外包公司编辑：程序博客网时间：2024/04/30 10:43

各种拙计，比赛时连暴力版本都没写出来,too weak :-(

参考代码：http://blog.csdn.net/kg_second/article/details/8034188

#include <cstdio>#include <cstring>using namespace std;const int maxn =1000+5;#define MAX 1000000000bool hash[maxn];int prime[maxn];int a[11];int n,k,cnt,K;void init(){    cnt=0;    memset(hash,0,sizeof(hash));    for(int i = 2 ;i < maxn;i ++)        if(!hash[i]){            prime[cnt++] = i;            for(int j= i+i ;j < maxn; j +=i)                hash[j] = 1;        }}void getprime()///素因子分解{    int tmp = n;    k = 0;    for(int i = 0;i < cnt&& prime[i] * prime[i] <= n; i++)        if(tmp%prime[i]==0){            a[k++] = prime[i];            while(tmp % prime[i] == 0)                tmp /= prime[i];        }    if(tmp != 1) a[k++] = tmp;}int judge(int num)///求出num前与num互质的元素个数{    int sum=0;    for(int i = 1; i < (1<<k); i++)///利用容斥原理    {        int tmp = 1,t = 0;        for(int j = 0;j < k; j++)            if((1 << j) & i){                t++;                tmp *= a[j];            }        if(t%2) sum += num/tmp;        else sum -= num/tmp;    }    return num-sum;}int main(){    init();    while(~scanf("%d%d",&n,&K))    {        getprime();        int l=1,r=MAX,mid,ans;        while(l<=r)        {            mid=(l+r)>>1;            int pt=judge(mid);            if(pt>=K){                if(pt==K) ans=mid;                r=mid-1;            }            else l=mid+1;        }        printf("%d\n",ans);    }    return 0;}