21.11.12模拟 P1445 [Violet]樱花 SP1434 KPEQU

POJ2917的加强版，把n变成了n!

n!的每个质因子数量就是1~n的对应质因子个数累加

这是模拟赛的，有取模的,P1445 [Violet]樱花

int prime[N], v[N];
	int cnt, c[N];

	inline void Euler_Sieve(int mx) {
		rep(i, 2, mx) {
			if(!v[i]) {
				v[i] = i;
				prime[++cnt] = i;
			}
			rep(j, 1, cnt) {
				if(prime[j] > v[i] || 1ll * prime[j]*i > mx) break;
				v[prime[j]*i] = prime[j];
			}
		}
	}

	int main() {
		Euler_Sieve(T);
		rep(i, 1, cnt) {
			int temp(T);
			while(1){
				if(temp==0||prime[i]>T) break;
				c[prime[i]]+=temp/prime[i];
				temp/=prime[i]; 
			}
		}
		lxl ans(1);
		rep(i, 1, T) {
			if(c[i]) {
				ans = (ans * (2 * c[i] + 1) ) % mod;
			}
		}
		out(ans, '
');
		return 0;
	}

这个是spoj的，要开高精度

const int N = 1000007;

const int len = 2607, W = 1e9;
struct Big {
	lxl x[len];
	int sz;
	void Trim() {
		while(sz > 1 && !x[sz - 1])--sz;
	}
	Big(lxl a = 0) {
		memset(x, 0, sizeof(x)), sz = 0;
		do x[sz++] = a % W;
		while(a /= W);
	}
} ans;
void Write(const Big & a) {
	printf("%lld", a.x[a.sz - 1]);
	drp(i, a.sz - 2, 0)printf("%09lld", a.x[i]);
}
Big operator*(const Big & a, const Big & b) {
	Big c;
	c.sz = a.sz + b.sz;
	rep(i, 0, a.sz - 1)rep(j, 0, b.sz - 1) {
		c.x[i + j] += a.x[i] * b.x[j];
		c.x[i + j + 1] += c.x[i + j] / W, c.x[i + j] %= W;
	}
	c.Trim();
	return c;
}


namespace solve {

	int prime[N], v[N];
	int cnt, c[N];

	inline void Euler_Sieve(int mx) {
		rep(i, 2, mx) {
			if(!v[i]) {
				v[i] = i;
				prime[++cnt] = i;
			}
			rep(j, 1, cnt) {
				if(prime[j] > v[i] || 1ll * prime[j]*i > mx) break;
				v[prime[j]*i] = prime[j];
			}
		}
	}

	int main() {
		T = 1e5;
		Euler_Sieve(T);
		while(1) {
			cin >> T;
			if(!T) break;
			memset(c,0,sizeof c);
			rep(i, 1, cnt) {
				int temp(T);
				while(1) {
					if(temp == 0 || prime[i] > T) break;
					c[prime[i]] += temp / prime[i];
					temp /= prime[i];
				}
			}
			ans = 1;
			rep(i, 1, T) {
				if(c[i]) {
					ans = ans * (2 * c[i] + 1) ;
				}
			}
			Write(ans);
			cout << '
';
		}
		return 0;
	}
}
main() {
	solve::main();
	return 0;
}