我们看这段代码
int cnt = 0;for (int a_1 = 0; a_1 <= m; a_1++) { for (int a_2 = 0; a_1 + a_2 <= m; a_2++) { ... for (int a_n = 0; a_1 + a_2 + ... + a_n <= m; a_n++) { cnt = (cnt + 1) % 19491001; } }}printf("%d\n", cnt);
其实是可以改写为
int cnt = 0;for (int a_1 = 1; a_1 <= m + n; a_1++) { for (int a_2 = 1; a_1 + a_2 <= m + n; a_2++) { ... for (int a_n = 1; a_1 + a_2 + ... + a_n <= m + n; a_n++) { cnt = (cnt + 1) % 19491001; } }}printf("%d\n", cnt);
答案不变(就是把\(a_0, a_1, ... , a_n\)全部加了1,源代码里相应的\(m\)要增加\(n\),因为n个循环变量,每个变量都增加了1,所需增加即为\(n \times 1 = n\))
然后根据组合数学中组合数的定义,所求为C(m + n, n)
由于数特~别~大~,而且19491001
是质数,所以这里使用了
哦对了还要用
下面代码
#include#define int long long#pragma GCC optimize(3)#pragma GCC optimize("Ofast")using namespace std;const int maxn = 20000000;const int p = 19491001LL;int n, inv[maxn], m, js[maxn];int Lucas(int n, int m){ if(n < m)return 0LL; if(n < p)return js[n] * inv[m] % p * inv[n - m] % p; return Lucas(n % p, m % p) * Lucas(n / p, m / p) % p;}signed main(){ int t; scanf("%lld", &t); js[0] = 1LL; for(register int i = 1LL; i <= p; i++)js[i] = js[i - 1] * i % p; inv[1] = 1LL; inv[0] = 1LL; for(register int i = 2LL; i <= p; i++)inv[i] = (p - p / i) * inv[p % i] % p; for(register int i = 2LL; i <= p; i++)inv[i] = inv[i] * inv[i - 1] % p; while(t--) { scanf("%lld%lld", &n, &m); printf("%lld\n", Lucas(n + m, m)); } return 0;}
三年OI一场空,不开long long见祖宗