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lightoj1035 欧拉函数(暴力)

2017-02-05 23:35 239 查看

题意

用表达式x=px11px22...的形式表示N!,1≤N≤100;

思路

先求出100以内的素数,然后暴力分解,记录每个素数出现的次数;

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2016
File Name   :欧拉函数暴力
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
#define lson rt << 1
#define rson rt << 1 | 1
#define bug cout << "BUG HERE\n"
#define debug(x) cout << #x << " = " << x << endl
#define ALL(v) (v).begin(), (v).end()
#define lowbit(x) ((x)&(-x))
#define Unique(x) sort(ALL(x)); (x).resize(unique(ALL(x)) - (x).begin())
#define BitOne(x) __builtin_popcount(x)
#define showtime printf("time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)
#define Rep(i, l, r) for (int i = l;i <= r;++i)
#define Rrep(i, r, l) for (int i = r;i >= l;--i)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
typedef pair<ii,int> iii;
const double eps = 1e-8;
const double pi = 4 * atan(1);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
int nCase = 0;
//精度正负、0的判断
int dcmp(double x){if (fabs(x) < eps) return 0;return x < 0?-1:1;}
template<class T> inline bool read(T &n){
T x = 0, tmp = 1;
char c = getchar();
while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
if(c == EOF) return false;
if(c == '-') c = getchar(), tmp = -1;
while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
n = x*tmp;
return true;
}
template <class T> inline void write(T n){
if(n < 0){putchar('-');n = -n;}
int len = 0,data[20];
while(n){data[len++] = n%10;n /= 10;}
if(!len) data[len++] = 0;
while(len--) putchar(data[len]+48);
}
LL QMOD(LL x, LL k) {
LL res = 1LL;
while(k) {if (k & 1) res = res * x % MOD;k >>= 1;x = x * x % MOD;}
return res;
}
int prime[100];
bool vis[100];
inline void init() {
for (int i = 2;i < 100;++i) {
if (!vis[i]) {
prime[++prime[0]] = i;
for (int j = i * i;j < 100;j += i)
vis[j] = true;
}
}
}
int a[100];
int main(int argc, const char * argv[])
{
// freopen("/Users/jamesqi/Desktop/in.txt","r",stdin);
// freopen("/Users/jamesqi/Desktop/out.txt","w",stdout);
// ios::sync_with_stdio(false);
// cout.sync_with_stdio(false);
// cin.sync_with_stdio(false);

int kase;cin >> kase;
init();
while(kase--) {
long long n;cin >> n;
memset(a, 0, sizeof a);
for (int i = 1;i <= n;++i) {
int x = i;
for (int j = 1;j <= prime[0] && prime[j] * prime[j] <= x;++j) {
if (x % prime[j] == 0) {
while(x % prime[j] == 0) {
++a[prime[j]];
x /= prime[j];
}
}
}
//very Important
if (x > 1) ++a[x];
}
vector<ii> buff;
for (int i = 2;i < 100;++i)
if (a[i] != 0) buff.push_back(ii(i, a[i]));
printf("Case %d: ", ++nCase);
printf("%lld =", n);
int Size = buff.size();
for (int i = 0;i < Size;++i) {
printf(" %d (%d)", buff[i].first, buff[i].second);
if (i != Size - 1) printf(" *");
}
printf("\n");
}

// showtime;
return 0;
}
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