方法： factorial mod， logarithm

?

求trailing zeros，其實就是factorial mod 的應用，

求長度，利用log 函數。需要注意的是，答案為int(log(n!)/log(b)) + 1, 比如 a = 2, b = 2, 長度為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 '\n'#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;} }

　　

（我的笨辦法）求trailing zero's 的長度，用一個數組記錄下factorial里各個prime的power，然后用另一個數組記錄下b里各個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 '\n'#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;} }

　　

轉載于:https://www.cnblogs.com/skyette/p/6357348.html

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