bzoj3944Sum

3944: Sum

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 5149  Solved: 1385
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Description

 

Input

一共T+1行

第1行为数据组数T(T<=10)

第2~T+1行每行一个非负整数N，代表一组询问

 

Output

一共T行，每行两个用空格分隔的数ans1,ans2

 

Sample Input

6
1
2
8
13
30
2333

Sample Output

1 1
2 0
22 -2
58 -3
278 -3
1655470 2

杜教筛入门？
http://blog.csdn.net/popoqqq/article/details/45023331

 #include<bits/stdc++.h>
 #define rint register int
 #define ll long long
 #define N 2000001
 using namespace std;
 int pri[N>>],vis[N],cnt,cas,n;
 ll phi[N],mo[N],p[N],q[N];
 void predeal(){
     mo[]=phi[]=;
     for(rint i=;i<N;++i){
         if(!vis[i]){phi[i]=i-;mo[i]=-;pri[++cnt]=i;}
         for(rint j=;j<=cnt&&pri[j]*i<N;++j){
             vis[i*pri[j]]=;
             if(i%pri[j]){phi[i*pri[j]]=phi[i]*(pri[j]-);mo[i*pri[j]]=-mo[i];}
             else{phi[i*pri[j]]=phi[i]*pri[j];mo[i*pri[j]]=;break;}
         }
     }
     for(rint i=;i<N;++i)phi[i]+=phi[i-],mo[i]+=mo[i-];
 }
 ll getp(int x){return x<N?phi[x]:p[n/x];}
 ll getq(int x){return x<N?mo[x]:q[n/x];}
 void solve(int x){
     if(x<N)return;int i,j=,t=n/x;
     if(vis[t])return;vis[t]=;
     p[t]=(1ll*x+)*x/;q[t]=;
     while(j<x){
         i=j+;j=x/(x/i);solve(x/i);
         p[t]-=getp(x/i)*(j-i+);
         q[t]-=getq(x/i)*(j-i+);
     }
 }
 int main(){
     scanf("%d",&cas);predeal();
     while(cas--){
         scanf("%d",&n);memset(vis,,sizeof(vis));
         if(n<N)printf("%lld %lld\n",phi[n],mo[n]);
         else{solve(n);printf("%lld %lld\n",p[],q[]);}
     }
     return ;
 }