LightOJ - 1090 Trailing Zeroes (II)

ⁿC_r * p^q

求该表达式末尾0的个数。

显然至于2 5 的因子个数有关。

考虑预处理出每个数的2，5的因子个数，并统计前缀和（题个的组合数其实是阶乘）

int _2[maxn], _5[maxn];
ll _s2[maxn], _s5[maxn];

int _count(int x, int y) {
    int res = 0;
    while (x % y == 0) x /= y, res++;
    return res;
}

void init() {
    for (int i = 2; i < maxn; i++) {
        _2[i] = _count(i, 2);
        _5[i] = _count(i, 5);
        _s2[i] = _s2[i - 1] + _2[i];
        _s5[i] = _s5[i - 1] + _5[i];
    }
}

int main() {
    init();
    int T = readint();
    int n, r, p, q;
    int kase = 1;
    while (T--) {
        n = readint(), r = readint(), p = readint(), q = readint();
        int c1 = _s2[n], c2 = _s5[n];
        c1 -= _s2[r] + _s2[n - r];
        c2 -= _s5[r] + _s5[n - r];
        int c3 = _2[p] * q;
        int c4 = _5[p] * q;
        printf("Case %d: %d\n", kase++, min((c1 + c3), (c2 + c4)));
    }
}