UVa 10061 How many zero's and how many digits?

方法： factorial mod， logarithm

求trailing zeros，其实就是factorial mod 的应用，

求长度，利用log 函数。需要注意的是，答案为int(log(n!)/log(b)) + 1, 比如 a = 2, b = 2, 长度为2.

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (int (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (int (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (int (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (int (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
  using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

int a, b;
bitset<2000001> vis(0);
ll primes[2000001];
int pcnt = 0;

// said to be O(n) prime generating
void init()
{
    for (ll i = 2; i <= 2e6; ++i)
    {
        if (!vis[i]) primes[pcnt++] = i;
        for (int j = 0; j < pcnt && i * primes[j] <= 2e6; ++j)
        {
            vis[j*primes[j]] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    init();

    while (cin >> a >> b)
    {

        int ans = 1e9;
        int B = b;
        for (int i = 0; i < pcnt && b >= primes[i]; ++i)
        {
            if (b % primes[i]) continue;
            int bcnt = 0;
            while (b % primes[i] == 0)
            {
                b /= primes[i];
                bcnt += 1;
            }
            int fcnt = 0;
            int cur = a;
            while (cur)
            {
                cur /= primes[i];
                fcnt += cur;
            }
            ans = min(ans, fcnt/bcnt);
        }
        double len = 0;
        for (int i = 2; i <= a; ++i)
        {
            len += log(1.0*i);
        }
        
        cout << ans << " " << max((int)(len/log(B)+1),1) << newline;
    }
}

　　

（我的笨办法）求trailing zero's 的长度，用一个数组记录下factorial里各个prime的power，然后用另一个数组记录下b里各个prime的power，然后求解。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (int (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (int (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (int (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (int (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
  using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

int a, b;
int d[2000001] = {0};
int cnt[2000001];
int fcnt[2000001];
void init()
{
    d[1] = 1;
    for (ll i = 2; i <= 2e6; ++i)
    {
        if (!d[i])
        {d[i] = (int)i;
        for (ll j = i*i; j <= 2e6; j += i)
            d[j] = (int) i;
        }
    }
}

int main()
{
    init();

    while (cin >> a >> b)
    {
        Reset(cnt, 0);
        int ans = 0;
        double len = 0;
        for (int i = 2; i <= a; ++i)
        {
            len += log(1.0*i);
            int cur = i;
            while (cur != 1)
            {
                cnt[d[cur]] += 1;
                cur /= d[cur];
            }
        }
        Reset(fcnt, 0);
        int cur = b;
        while (cur != 1)
        {
            fcnt[d[cur]] += 1;
            cur /= d[cur];
        }
        ans = 1e9;
        for (int i = 2; i <= b; ++i)
            if (fcnt[i])
            {
                ans = min(ans, cnt[i]/fcnt[i]);
                //cerr << i << " " << cnt[i] << " " << fcnt[i] << newline;
            }
        cout << ans << " " << max((int)(len/log(b)+1),1) << newline;
    }
}