hdu5750Dertouzos

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链接：http://acm.hdu.edu.cn/showproblem.php?pid=5750

题意：中文题意。

分析：用类似均摊的方法在遍历素数的时候加上n/d这个限制会优化。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=40010;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=1000000007;const int MAX=2000000010;const ll INF=1ll<<55;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;int k,a[N],q[N],f[N];void deal(int n) {    int i,j;k=0;    for (i=2;i<=n;i++) {        if (!q[i]) a[++k]=i;        for (j=1;j<=k;j++) {            if (a[j]*i>n) break ;            q[a[j]*i]=1;            if (i%a[j]==0) break ;        }    }}int main(){    int i,t,l,r,mid;    ll n,d,g;    scanf("%d", &t);    deal(40000);    while (t--) {        scanf("%I64d%I64d", &n, &d);        if (d>=n||d*2ll>=n) printf("0\n");        else {            for (i=1;i<=3450&&a[i]<=n/d;i++)            if (a[i]>d||d%a[i]==0) break ;            if (a[i]>d) g=1ll*a[i-1];            else g=1ll*a[i];            l=1;r=k;mid=(l+r)>>1;            while (l+1<r)            if (1ll*a[mid]>g||1ll*a[mid]*d>=n) { r=mid;mid=(l+r)>>1; }            else { l=mid;mid=(l+r)>>1; }            printf("%d\n", l);        }    }    return 0;}

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