LA 7362 Farey (数学，欧拉函数)

题意：给定一个数 n，问你0<= a <=n, 0 <= b <= n，有多少个不同的最简分数。

析：这是一个欧拉函数题，由于当时背不过模板，又不让看书，我就暴力了一下，竟然AC了，才2s，题目是给了3s，很明显是由前面递推，前面成立的，后面的也成立，

只要判定第 i 个有几个，再加前 i-1 个就好，第 i 个就是判断与第 i 个互质的数有多少，这就是欧拉函数了。

代码如下：

这是欧拉函数的。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10000 + 5;
const int mod = 1e9;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int ans[maxn];
int phi[maxn];

void init(){
    memset(phi, 0, sizeof(phi));
    phi[1] = 1;
    for(int i = 2; i <= 10000; ++i) if(!phi[i])
        for(int j = i; j <= 10000; j += i){
            if(!phi[j])  phi[j] = j;
            phi[j] = phi[j] / i * (i-1);
        }

    ans[2] = 3;
    for(int i = 3; i <= 10000; ++i)
        ans[i] = ans[i-1] + phi[i];
}

int main(){
    init();
    int T;   cin >> T;
    while(T--){
        scanf("%d %d", &m, &n);
        printf("%d %d\n", m, ans[n]);
    }
    return 0;
}

这是我暴力的：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const int mod = 1e9;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int ans[10005];

int main(){
    ans[1] = 2;  ans[2] = 3;
    for(int i = 3; i <= 10000; ++i){
        int cnt = 0;
        for(int j = 1; j <= i/2; ++j){
            if(__gcd(j, i) == 1) ++cnt;
        }
        ans[i] = ans[i-1] + 2*cnt;
    }
    int T; cin >> T;
    while(T--){
        scanf("%d %d", &m, &n);
        printf("%d %d\n", m, ans[n]);
    }
    return 0;
}