【LightOJ】1370 - Bi-shoe and Phi-shoe（欧拉函数，素数打表）
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1370 - Bi-shoe and Phi-shoe
Bamboo Pole-vault is a massively popular sport in Xzhiland. And Master Phi-shoe is a very popular coach for his success. He needs some bamboos for his students,so he asked his assistant Bi-Shoe to go to the market and buy them. Plenty of Bamboos of all possible integer lengths (yes!) are available in the market. According to Xzhila tradition, Score of a bamboo = Φ (bamboo's length) (Xzhilans are really fond of number theory). For your information,Φ (n) = numbers less than n which are relatively prime (having no common divisor other than 1) to n. So,score of a bamboo of length 9 is 6 as 1,2,4,5,7,8 are relatively prime to 9. The assistant Bi-shoe has to buy one bamboo for each student. As a twist,each pole-vault student of Phi-shoe has a lucky number. Bi-shoe wants to buy bamboos such that each of them gets a bamboo with a score greater than or equal to his/her lucky number. Bi-shoe wants to minimize the total amount of money spent for buying the bamboos. One unit of bamboo costs 1 Xukha. Help him. InputInput starts with an integer T (≤ 100),denoting the number of test cases. Each case starts with a line containing an integer n (1 ≤ n ≤ 10000) denoting the number of students of Phi-shoe. The next line contains n space separated integers denoting the lucky numbers for the students. Each lucky number will lie in the range [1,106]. OutputFor each case,print the case number and the minimum possible money spent for buying the bamboos. See the samples for details.
? 这里用欧拉函数，即：一个素数的比它小的互质数的个数为该数减一。这道题之用了这一点。 但是这里还要注意一个知识点，假如ψ(N)的欧拉函数值为p，则N的最小值为大于N的最小素数。 代码如下： #include <cstdio>
int su[1000111]={1,1};
int main()
{
for (int i=2;i<=1000111;i++) //打素数表，素数为0
{
if (su[i]==1)
continue;
for (int j=i+i;j<=1000111;j+=i)
su[j]=1;
}
int u;
int n;
long long ans;
int num=1;
scanf ("%d",&u);
while (u--)
{
scanf ("%d",&n);
ans=0;
while (n--)
{
int t;
scanf ("%d",&t);
for (int i=t+1;;i++)
{
if (su[i]==0)
{
ans+=i;
break;
}
}
}
printf ("Case %d: %lld Xukhan",num++,ans);
}
return 0;
}
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