How hdu 1695 hdu 4135 is used to calculate the number of mutual quality with r in the 1-n range.

Method:

First, the prime factor of n is decomposed, and each prime factor is recorded separately. Then the number of non-interconnectivity with a certain prime factor in the obtained range is n/r (I). Assume that r (I) is a qualitative factor of r.

Assuming there are only three quality factors, the total number of non-interphysical should be p1 + p2 + p3-p1 * P2-P1 * p3-p2 * p3 + p1 * p2 * p3, and the principle of rejection, you can go to Baidu encyclopedia to view related content

Pi represents n/r (I), that is, the number of numbers that do not interact with a certain quality factor.

When there are more quality factors, they can be solved by State compression. If the binary bit is 1, this quality factor is obtained. If there is an odd number of 1, it is added, and vice versa.

View Code

#include<vector>
#include<cstdio>
using namespace std;
__int64 solve(int r,__int64 n){
    vector<int> p;
int i;
for(i=2;i*i<=r;i++){
if(r%i==0){
            p.push_back(i);
while(r%i==0){
                r/=i;
            }
        }
    }
if(r>1) p.push_back(r);
    __int64 sum=0;
for(int num=1;num<(1<<p.size());num++){
        __int64 mult=1,ones=0;
for(i=0;i<p.size();i++){
if(num&(1<<i)){
                ones++;
                mult*=p[i];
            }
        }
if(ones%2) sum+=n/mult;
else sum-=n/mult;
    }
return n-sum;
}
int main(){
int t,c;
    __int64 a,b;
    __int64 cases=1;
    scanf("%d",&t);
while(t--){
        scanf("%I64d%I64d%d",&a,&b,&c);
        printf("Case #%I64d: %I64d\n",cases++,solve(c,b)-solve(c,a-1));
    }
return 0;
}

Hdu 1695

View Code

#include<vector>#include<cstdio>using namespace std;int prm[50000];int tot=0;bool flag[100010];void init(){    int i,j;    memset(flag,false,sizeof(flag));    for(i=2;i<=100000;i++)    {        for(j=2*i;j<=100000;j+=i)        {            flag[j]=true;        }    }    for(i=2;i<=100000;i++)    {        if(!flag[i])            prm[++tot]=i;    }}__int64 solve(int r,__int64 n){    int p[50];    int cnt=0;    int i;    if(!flag[r])    {        if(r>1)            p[cnt++]=r;    }    else     {        for(i=1;i<=tot;i++)        {            if(r%prm[i]==0)            {                p[cnt++]=prm[i];                while(r%prm[i]==0)        r/=prm[i];                if(!flag[r])                {                    if(r>1)                    {                        p[cnt++]=r;r=1;                    }                    break;                }            }        }if(r>1) p[cnt++]=r;    }    __int64 sum=0;    for(int num=1;num<(1<<cnt);num++){        __int64 mult=1,ones=0;        for(i=0;i<cnt;i++){            if(num&(1<<i)){                ones++;                mult*=p[i];            }        }        if(ones%2) sum+=n/mult;        else sum-=n/mult;    }    return n-sum;}int main(){    int t;    init();    __int64 a,b,c,d,e;    __int64 cases=1;    scanf("%d",&t);    while(t--){        scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&e);        if(e==0||b<e||d<e)         {            printf("Case %I64d: 0\n",cases++);            continue;        }         a = b < d ? b : d;                 b = b > d ? b : d;                 a /= e;                 b /= e;        __int64 ans=0;        for(int i=1;i<=a;i++)        {            if(i>1)            ans+=solve(i,b)-solve(i,i-1);            else ans+=b;        }        printf("Case %I64d: %I64d\n",cases++,ans);    }    return 0;}