UVALive 7362 Farey （欧拉函数）

大体题意：

告诉你n (n＜=10000)  求解有多少个分数 满足分母  1 < b <= n   分子  a 与b 互质？

思路：

这种问法很显然的是欧拉函数！

定义：

欧拉函数f(n) 表示小于或等于n 的书中与n 互质的数的数目。

(1).f(1) = 1;

(2).若n 是素数p 的k次幂，f(n) = p^k - p^k-1 = (p-1)*p^(k-1)

(3).如果m 和n 互质，f(m*n) = f(m) * f(n);

所以整体思路就有了，直接枚举b  对b用欧拉函数即可！

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<cctype>
#include<string>
#include<iostream>
#include<queue>
#define fout freopen("out.txt","w","stdout");
#define mr make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef unsigned long long LLU;
const int maxn = 1000 + 5;
const int mod = 1000000000;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
int gcd(int a,int b){
return !b?a:gcd(b,a%b);
}
int dp[10007];
int vis[10007];
vector<int>prime;
int len_prime;
void init(){
dp[1] = 2;
int len = sqrt(10000) + 1;
for (int i = 2; i <= len; ++i) if (!vis[i]){
for (int j = i*i; j < 10000; j+=i)vis[j] = 1;

}
for (int i = 2; i < 10000; ++i)if (!vis[i])prime.push_back(i);
len_prime=(int)prime.size();
for (int i = 2; i <= 10000; ++i){
int ans = i;
int n = i;
for (int j = 0; j < len_prime; ++j){
int pri = prime[j];
if (n % pri == 0){
ans = ans/pri * (pri-1);
while(n % pri == 0)n/=pri;
}
}
//        printf("ans = %d\n",ans);
dp[i] = dp[i-1] + ans;
}
}
int main(){

int T;
init();
scanf("%d",&T);
while(T--){
ll k,n;
scanf("%lld %lld",&k,&n);
printf("%lld %d\n",k,dp
);
}
return 0;
}