Longge的问题(欧拉,思维)
Description
Input
Output
Sample Input
6
Sample Output
15
Hint
【数据范围】
对于60%的数据,0<N<=2^16。
对于100%的数据,0<N<=2^32。
题解:题目让求∑gcd(i, N)(1<=i <=N);很显然暴力是不行的,由于gcd(m,n)=k,所以gcd(m/k,n/k)=1;要求小于等于n的最大公约数是k的个数num[k]等于ouler[n/k];枚举k求k*num[k]的和即可;
代码:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<vector>
#include<map>
#include<string>
using namespace std;
typedef long long LL;
LL ouler(LL n){
LL ans=n;
for(LL i=;i*i<=n;i++){
if(n%i==){
ans=ans*(i-)/i;
while(n%i==)n/=i;
}
}
if(n>)ans=ans*(n-)/n;
return ans;
}
int main(){
LL N;
while(~scanf("%lld",&N)){
LL ans=;
for(LL i=;i*i<=N;i++){
if(N%i==){
ans+=ouler(N/i)*i;
if(N/i>sqrt(1.0*N))ans+=ouler(i)*(N/i);//这里ouler[N/N/i]=ouler[i]计算最大公约数是N/i的个数
}
}
printf("%lld\n",ans);
}
return ;
}
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