多校第四场 1003 HDU 6069 素数晒应用

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<div class="dp-highlighter bg_cpp"><div class="bar"><div class="tools">[cpp] <a href="#" class="ViewSource" title="view plain" onclick="dp.sh.Toolbar.Command('ViewSource',this);return false;" target="_blank">view plain</a> <a href="#" class="CopyToClipboard" title="copy" onclick="dp.sh.Toolbar.Command('CopyToClipboard',this);return false;" target="_blank">copy</a><div style="position: absolute; left: 455px; top: 472px; width: 18px; height: 18px; z-index: 99;"><embed id="ZeroClipboardMovie_1" src="http://static.blog.csdn.net/scripts/ZeroClipboard/ZeroClipboard.swf" loop="false" menu="false" quality="best" bgcolor="#ffffff" width="18" height="18" name="ZeroClipboardMovie_1" align="middle" allowscriptaccess="always" allowfullscreen="false" type="application/x-shockwave-flash" pluginspage="http://www.macromedia.com/go/getflashplayer" flashvars="id=1&width=18&height=18" wmode="transparent"></div> <a href="#" class="PrintSource" title="print" onclick="dp.sh.Toolbar.Command('PrintSource',this);return false;" target="_blank">print</a><a href="#" class="About" title="?" onclick="dp.sh.Toolbar.Command('About',this);return false;" target="_blank">?</a></div></div><ol start="1" class="dp-cpp"><li class="alt">#include<iostream>  </li><li class="">#include<queue>  </li><li class="alt">#include<cstring>  </li><li class="">#include<vector>  </li><li class="alt">#include<string>  </li><li class="">#include<algorithm>  </li><li class="alt">#include<cstdio>  </li><li class="">#include<cmath>  </li><li class="alt">#include<set>  </li><li class="">#include<cstdlib>  </li><li class="alt">#include<stack>  </li><li class="">typedef long long LL;  </li><li class="alt">const int maxsize = 1e6+7;  </li><li class="">using namespace std;  </li><li class="alt">int Ans[maxsize];  </li><li class="">LL D[maxsize];  </li><li class="alt">const int mood = 998244353;  </li><li class="">bool Prime[maxsize];  </li><li class="alt">LL PrimeT[maxsize];  </li><li class="">int pcnt;  </li><li class="alt">void primeFun(int n)//筛选法构造素数  </li><li class="">{  </li><li class="alt">    pcnt = 0;  </li><li class="">    Prime[0] = Prime[1] = 1;  </li><li class="alt">    for(int i=2;i*i<=n+1;i++)  </li><li class="">    {  </li><li class="alt">        if(!Prime[i])  </li><li class="">        {  </li><li class="alt">            for(int j=i+i;j<=n;j+=i)  </li><li class="">            {  </li><li class="alt">                Prime[j] = 1;  </li><li class="">            }  </li><li class="alt">        }  </li><li class="">    }  </li><li class="alt">    for(int i=0;i<=n;i++)  </li><li class="">    {  </li><li class="alt">        if(!Prime[i])  </li><li class="">        {  </li><li class="alt">            PrimeT[pcnt++] = i;  </li><li class="">            //cout<<i<<" ";  </li><li class="alt">        }  </li><li class="">    }  </li><li class="alt">}  </li><li class="">  </li><li class="alt">  </li><li class="">int main()  </li><li class="alt">{  </li><li class="">    //freopen("finput.txt","r",stdin);  </li><li class="alt">    primeFun(1e6+1);  </li><li class="">    int t;  </li><li class="alt">    scanf("%d",&t);  </li><li class="">    while(t--)  </li><li class="alt">    {  </li><li class="">        LL l,r,k;  </li><li class="alt">        scanf("%lld%lld%lld",&l,&r,&k);  </li><li class="">        LL n = r-l+1;  </li><li class="alt">        fill(Ans,Ans+n,1);//赋值为1  </li><li class="">        for(int i=0;i<n;i++)  </li><li class="alt">        {  </li><li class="">            D[i] = l+i;  </li><li class="alt">        }  </li><li class="">        for(int i=0;i<pcnt;i++)  </li><li class="alt">        {  </li><li class="">            LL primenow = PrimeT[i];  </li><li class="alt">            LL t = l/primenow;  </li><li class="">            if(l%primenow) t++;  </li><li class="alt">            t *= primenow;  </li><li class="">            for(LL j=t-l;j<n;j+=primenow)//算法的关键所在,应为是跳着取得,所以只有prime是这个数的因子的时候才进行操作  </li><li class="alt">            {  </li><li class="">                int cnt = 0;  </li><li class="alt">                while(D[j]%primenow==0)  </li><li class="">                {  </li><li class="alt">                    cnt++;  </li><li class="">                    D[j]/=primenow;  </li><li class="alt">                }  </li><li class="">                Ans[j] = (Ans[j]*(cnt*k+1))%mood;  </li><li class="alt">            }  </li><li class="">        }  </li><li class="alt">        LL sum = 0;  </li><li class="">        for(int i=0;i<n;i++)  </li><li class="alt">        {  </li><li class="">            if(D[i]!=1)//对于大于10^6的整数进行特殊处理  </li><li class="alt">            {  </li><li class="">                Ans[i] = (Ans[i]*(k+1))%mood;  </li><li class="alt">            }  </li><li class="">            //cout<<Ans[i]<<" ";  </li><li class="alt">            sum = (sum+Ans[i])%mood;  </li><li class="">        }  </li><li class="alt">        printf("%lld\n",sum);  </li><li class="">    }  </li><li class="alt">    return 0;  </li><li class="">}  </li></ol></div><pre code_snippet_id="2516235" snippet_file_name="blog_20170803_1_2386237" name="code" class="cpp" style="display: none;"> #include<iostream>
#include<queue>
#include<cstring>
#include<vector>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<set>
#include<cstdlib>
#include<stack>
typedef long long LL;
const int maxsize = 1e6+7;
using namespace std;
int Ans[maxsize];
LL D[maxsize];
const int mood = 998244353;
bool Prime[maxsize];
LL PrimeT[maxsize];
int pcnt;
void primeFun(int n)//筛选法构造素数
{
pcnt = 0;
Prime[0] = Prime[1] = 1;
for(int i=2;i*i<=n+1;i++)
{
if(!Prime[i])
{
for(int j=i+i;j<=n;j+=i)
{
Prime[j] = 1;
}
}
}
for(int i=0;i<=n;i++)
{
if(!Prime[i])
{
PrimeT[pcnt++] = i;
//cout<<i<<" ";
}
}
}

int main()
{
//freopen("finput.txt","r",stdin);
primeFun(1e6+1);
int t;
scanf("%d",&t);
while(t--)
{
LL l,r,k;
scanf("%lld%lld%lld",&l,&r,&k);
LL n = r-l+1;
fill(Ans,Ans+n,1);//赋值为1
for(int i=0;i<n;i++)
{
D[i] = l+i;
}
for(int i=0;i<pcnt;i++)
{
LL primenow = PrimeT[i];
LL t = l/primenow;
if(l%primenow) t++;
t *= primenow;
for(LL j=t-l;j<n;j+=primenow)//算法的关键所在,应为是跳着取得,所以只有prime是这个数的因子的时候才进行操作
{
int cnt = 0;
while(D[j]%primenow==0)
{
cnt++;
D[j]/=primenow;
}
Ans[j] = (Ans[j]*(cnt*k+1))%mood;
}
}
LL sum = 0;
for(int i=0;i<n;i++)
{
if(D[i]!=1)//对于大于10^6的整数进行特殊处理
{
Ans[i] = (Ans[i]*(k+1))%mood;
}
//cout<<Ans[i]<<" ";
sum = (sum+Ans[i])%mood;
}
printf("%lld\n",sum);
}
return 0;
}
</pre> 
 

 

 


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