[Dwango Programming Contest 6th C] Cookie Distribution
Posted suwakow
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[Dwango Programming Contest 6th C] Cookie Distribution相关的知识,希望对你有一定的参考价值。
组合意义,妙啊(弱菜 swk 不会数数实锤了
考虑 (prod c_i) 的组合意义,它代表每个孩子在他得到的饼干中选择一个拿出来,最终的方案数。
因此得到了一个几乎显然的 dp:
设 (f_{i, j}) 为前 (i) 天,有 (j) 个孩子已经 “选择了” 他的饼干的方案数。转移时枚举新增了 (u) 个孩子,则转移系数是一系列组合数,详见代码。
#include <bits/stdc++.h>
#define R register
#define mp make_pair
#define ll long long
#define pii pair<int, int>
using namespace std;
const int N = 1100, K = 22, mod = 1e9 + 7;
int n, k, a[K];
ll f[K][N], comb[N][N];
template <class T>
inline void read(T &x) {
x = 0;
char ch = getchar(), w = 0;
while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
x = w ? -x : x;
return;
}
inline ll addMod(ll a, ll b) {
return (a += b) >= mod ? a - mod : a;
}
int main() {
read(n), read(k);
for (R int i = 1; i <= k; ++i) read(a[i]);
for (R int i = 0; i <= n; ++i) {
comb[i][0] = 1;
for (R int j = 1; j <= i; ++j)
comb[i][j] = addMod(comb[i - 1][j - 1], comb[i - 1][j]);
}
f[0][0] = 1;
for (R int i = 1; i <= k; ++i)
for (R int j = 0; j <= n; ++j)
for (R int l = max(0, j - a[i]); l <= j; ++l)
f[i][j] = (f[i][j] + f[i - 1][l] * comb[n - l][j - l] % mod * comb[n - j + l][a[i] - j + l]) % mod;
cout << f[k][n] << endl;
return 0;
}
以上是关于[Dwango Programming Contest 6th C] Cookie Distribution的主要内容,如果未能解决你的问题,请参考以下文章
Dwango Programming Contest 6th C
Dwango Programming Contest 6th Task C. Cookie Distribution
[Dwango Programming Contest 6th C] Cookie Distribution
English Programming Lesson: Vocabulary about Programming