2015东北四省赛 L题 线性筛+积性函数 求因子和

Given an interval [x, y], calculate the sum of all the factors of the whole integers from x to y (include x and y).

Let describe it officially. Let's definite f (i) is the sum of factors of i, you should work out S(x, y), here S(x, y) is:

输入

There are multiple test cases, the number of cases T (1 ≤ T ≤ 105) will be given in the first line of input data. 
Each test case will include two integer x and y (1 ≤ x < y ≤ 5 × 106).

输出

For the ith test case, output “Case $i:” in the first line.
And print the answer of S(x, y) in a single line. The answer may be very large, so moudule 10000007 before printing it.

样例输入

31 65 1212 300

样例输出

Case $1:33Case $2:112Case $3:74210

提示

F (1) = 1 = 1


f (2) = 1 + 2 = 3


f (3) = 1 + 3 = 4


f (4) = 1 + 2 + 4 = 7


f (5) = 1 + 5 = 6


f (6) = 1 + 2 + 3 + 6 = 12


So the answer is 1 + 3 + 4 + 7 + 6 + 12 = 33

来源

2015大连海事

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=5000000+10;
const int mod=10000007;
int vis[maxn];
ll prime[maxn];
ll fac[maxn];//因子和积性函数
int num[maxn];//i这个数的最小质因子的幂指数
int qpow(int a,int b)
{
int res=1;
while(b)
{
if(b&1)
res=res*a;
a=a*a;
b=b/2;
}
return res;
}
void init()
{
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
int cnt=0;
fac[1]=1;
fac[0]=0;
for(ll i=2;i<=maxn;i++)
{
if(!vis[i])
{
prime[cnt++]=i;
fac[i]=1+i;
num[i]=1;
}
for(ll j=0;j<cnt&&i*prime[j]<=maxn;j++)
{
vis[i*prime[j]]=1;//该数的最小质因子为prime[j]
if(i%prime[j]==0)//若这个数为2*6 或者为2*4（2^3）
{
num[i*prime[j]]=num[i]+1;
//判断这个数是否为p^k次方形式
int res=i;
while(res%prime[j]==0)
{res=res/prime[j];}
int ok=1;
if(res==1)
ok=0;
//
if(ok==0)//若为p^k次方形式,则需单独计算 f[p^k]=f[p^k-1]+p^k;
fac[i*prime[j]]=fac[i]+qpow(prime[j],num[i]+1);
else//若不为，则f
=f[p^k]*f[n/(p^k)];
fac[i*prime[j]]=fac[i/qpow(prime[j],num[i])]*fac[qpow(prime[j],num[i]+1)];
break;
}
else//若这个数为互质的数相乘
{
num[i*prime[j]]=1;
fac[i*prime[j]]=fac[i]*fac[prime[j]];
}
}
}
// for(int i=1;i<=12;i++)
//    printf("%d ",fac[i]);
for(int i=2;i<=maxn;i++)
fac[i]=(fac[i-1]+fac[i])%mod;

}
int main()
{
init();
int t;
int cas=1;
cin>>t;
while(t--)
{
int x,y;
scanf("%d%d",&x,&y);
printf("Case $%d:\n",cas++);
printf("%lld\n",(fac[y]-fac[x-1]+mod)%mod);
}

}