【欧拉函数】 TOJ 4125. Game

从大到小枚举约数。。。然后就是欧拉函数就是当前约数的个数。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 1000005
#define maxm 5000005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL; 
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

LL a[maxn];
LL b[maxn];
LL aa[maxn];
LL bb[maxn];
LL phi[maxn];

void init()
{
	int n = 1000000;
	phi[1] = 1;
	for(int i = 2; i <= n; i++) if(!phi[i]) 
		for(int j = i; j <= n; j += i) {
			if(!phi[j]) phi[j] = j;
			phi[j] -= phi[j] / i;
		}
	for(int i = n; i >= 1; i--)
		for(int j = i; j <= n; j += i) {
			int t = phi[j / i];
			if(aa[j] == bb[j]) {
				a[j] += (LL)i * ((t+1) / 2);
				b[j] += (LL)i * (t / 2);
				aa[j] += (t+1) / 2;	
				bb[j] += t / 2;
			}
			else {
				a[j] += (LL)i * (t / 2);
				b[j] += (LL)i * ((t+1) / 2);
				aa[j] += t / 2;
				bb[j] += (t+1) / 2;
			}
		}
}

int main()
{
	init();
	int n;
	while(scanf("%d", &n) != EOF) printf("%lld %lld\n", a[n], b[n]);
	
	return 0;
}


你可能感兴趣的:(数论,欧拉函数)