[LightOJ1028]Trailing Zeroes (I)（拆分因子）

题目链接：http://lightoj.com/volume_showproblem.php?problem=1028

思路：求一个数的所有因子数，结果-1…数据比较大，所以要加一步优化t < prime[i] * prime[i]的时候要退出。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 1001000;
LL n;
LL prime[maxn], vis[maxn];
int pcnt;

void init() {
    pcnt = 0; Cls(vis);
    for(LL i = 2; i < maxn; i++) {
        if(!vis[i]) {
            for(LL j = i * i; j < maxn; j+=i)
            vis[j] = 1;
            prime[pcnt++] = i;
        }
    }
}

LL factor(LL x) {
    LL ret = 1;
    LL t = x;
    Rep(i, pcnt) {
        if(t < prime[i] * prime[i]) break;
        if(t % prime[i] == 0) {
            int k = 0;
            while(t % prime[i] == 0) {
                t /= prime[i];
                k++;
            }
            ret *= (k + 1);
        }
    }
    if(t > 1) ret *= 2;
    return ret;
}

int main() {
    // FRead();
    init();
    int T, _ = 1;
    Rint(T);
    W(T) {
        printf("Case %d: ", _++);
        cin >> n;
        cout << factor(n) - 1 << endl;
    }
}