标签:sizeof ring problem 情况 clu file ini == math
嘟嘟嘟
此题不难。
这种题做几道就知道些套路了:我们枚举酒有几堆,这样就能算出食物有多少堆以及他们的排列数,那么概率就是合法方案数 / 总方案数。
设酒有\(i\)堆,那么就有\(C_{w - 1} ^ {i - 1}\)种排列方法,对应的食物堆数就可能有\(i - 1, i, i + 1\)堆,然后同样用隔板法算出食物的排列方法,即\(C_{f - 1} ^ {i - 2}, C_{f - 1} ^ {i - 1}, C_{f - 1} ^ {i}\)。把这俩乘起来就是当酒堆数为\(i\)的总方案数。
至于合法方案数,就是我们先强制往每一堆酒上放\(h\)个,然后再往\(i\)堆酒上放\(w - h * i\)个,即\(C_{w - h * i - 1} ^ {i - 1}\)。
然后注意边界情况。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<assert.h>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
const ll mod = 1e9 + 7;
In ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
In void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
freopen(".in", "r", stdin);
freopen(".out", "w", stdout);
#endif
}
int n, m, h;
ll fac[maxn], inv[maxn];
In ll inc(ll a, ll b) {return a + b < mod ? a + b : a + b - mod;}
In ll C(int n, int m)
{
if(m < 0 || m > n) return 0;
return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
In ll quickpow(ll a, ll b)
{
ll ret = 1;
for(; b; b >>= 1, a = a * a % mod)
if(b & 1) ret = ret * a % mod;
return ret;
}
In void init()
{
fac[0] = inv[0] = 1;
for(int i = 1; i < maxn; ++i) fac[i] = fac[i - 1] * i % mod;
inv[maxn - 1] = quickpow(fac[maxn - 1], mod - 2);
for(int i = maxn - 2; i; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
}
int main()
{
//MYFILE();
m = read(), n = read(), h = read();
if(!m) {puts(n > h ? "1" : "0"); return 0;}
if(!n) {puts("1"); return 0;}
init();
ll ans1 = 0, ans2 = 0;
for(int i = 1; i <= n; ++i)
{
ll tp1 = C(n - 1, i - 1);
ll tp2 = inc(C(m - 1, i - 2), inc(C(m - 1, i - 1) * 2, C(m - 1, i)));
if(1LL * n - 1LL * h * i >= i) ans2 = inc(ans2, C(n - h * i - 1, i - 1) * tp2 % mod);
ans1 = inc(ans1, tp1 * tp2 % mod);
}
write(ans2 * quickpow(ans1, mod - 2) % mod), enter;
return 0;
}
标签:sizeof ring problem 情况 clu file ini == math
原文地址:https://www.cnblogs.com/mrclr/p/10935034.html