CF768F. Barrels and boxes

Solution

又一道据说是套路题。

核心在于枚举$W$的连续段个数为$x$。
然后就能知道$F$的连续段个数为$x-1,x,x+1$，可以用插板法求出任意把$W$分$x$段的方案数，分$x$段每段$W$至少$H$个的方案数，以及任意把$F$分$x-1,x,x+1$段的方案数了。

（注意一些边界情况）

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+7;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int n,fac[MAXN],inv[MAXN];
int quick_pow(int x,int y)
{int ret=1;for (;y;y>>=1){if (y&1) ret=1ll*ret*x%mods;x=1ll*x*x%mods;}return ret;
}
void Init(int n)
{fac[0]=1;for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mods;inv[n]=quick_pow(fac[n],mods-2);for (int i=n-1;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%mods;
}
ll C(int x,int y) { if (x<y||y<0) return 0; return 1ll*fac[x]*inv[y]%mods*inv[x-y]%mods; }
int upd(int x,int y) { return x+y>=mods?x+y-mods:x+y; }
signed main()
{int f=read(),w=read(),h=read(),ans1=0,ans2=0;if (!f) { puts(w<=h?"0":"1"); return 0; }if (!w) { puts("1"); return 0; }Init(f+w);for (int i=1;i<=w;i++) ans1=upd(ans1,C(w-1,i-1)*(C(f-1,i-2)+C(f-1,i-1)+C(f-1,i-1)+C(f-1,i))%mods);for (int i=1;i<=w&&w-h*i-1>=0;i++) ans2=upd(ans2,C(w-h*i-1,i-1)*(C(f-1,i-2)+C(f-1,i-1)+C(f-1,i-1)+C(f-1,i))%mods);printf("%lld\n",1ll*ans2*quick_pow(ans1,mods-2)%mods);return 0;
}