ABC-405-E - Fruit Lineup · znzryb
## 思路讲解
像这种题目，一般要把一对有关系的留到最后，这样子状态就唯一了。
```cpp
|----A----| |--D--|
|-------B------|
|-----C----|
```
可以先枚举A的边界，然后再把C填好，B，D之间因为有位置关系，所以状态唯一。
```cpp
// 显然是要枚举水果的边界
ll ans=0;
FOR(i,a,a+b){
// 在i-1个空位中选a-1个空位(因为第i个位置已经钦定了A（苹果）)
ll res=comb.C(i-1,a-1);
res*=comb.C(N-i,c);
res%=mod;
ans+=res;
}
```
## AC代码
[https://atcoder.jp/contests/abc405/submissions/66027443](https://atcoder.jp/contests/abc405/submissions/66027443)
::: 源代码
```cpp
// Problem: E - Fruit Lineup
// Contest: AtCoder - AtCoder Beginner Contest 405
// URL: https://atcoder.jp/contests/abc405/tasks/abc405_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
do { \
std::cerr << #variable << ":"; \
for (const auto& elem : variable) { \
std::cerr << elem << " "; \
} \
std::cerr << "\n"; \
} while (0)
#define uniVec(var) \
do { \
sort(var.begin(),var.end());\
var.resize(unique(var.begin(),var.end())-var.begin());\
} while (0)
#define debugN(var,N) \
do{ \
std::cerr<<#var<<":"; \
FOR(i,1,N){ \
std::cerr<<var[i]<<" "; \
} \
std::cerr<<"\n"; \
}while(0)
#define debugMap(variable) \
do { \
std::cerr << #variable << ":\n"; \
for (const auto& pair : variable) { \
std::cerr << " " << pair.first << " => " << pair.second << "\n"; \
} \
std::cerr << std::endl; \
} while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=998244353;
constexpr double eps=1e-8;
ll N,M,K,T;
/*
*/
struct Comb{
vector<ll> frac,infrac;
int ln=1;
Comb(int n){
ln=n;
frac.resize(n+5,1);
infrac.resize(n+5,1);
for(int i=1;i<=ln;++i){
frac[i]=frac[i-1]*i;
frac[i]%=mod;
}
for(int i=1;i<=ln;++i){
infrac[i]=binpow(frac[i],mod-2);
}
}
inline ll binpow(ll a,ll k){
ll res=1;
while(k){
if(k&1) res*=a;
a*=a;
a%=mod;
res%=mod;
k>>=1;
}
return res;
}
inline ll C(ll n,ll k){
ll res=frac[n]*infrac[n-k];
res%=mod;
res*=infrac[k];
res%=mod;
return res;
}
inline ll P(ll n,ll k){
ll res=frac[n]*infrac[n-k];
res%=mod;
return res;
}
};
inline void solve(){
ll a,b,c,d;
cin>>a>>b>>c>>d;
N=a+b+c+d;
Comb comb(a+b+c+d);
// 显然是要枚举水果的边界
ll ans=0;
FOR(i,a,a+b){
// 在i-1个空位中选a-1个空位(因为第i个位置已经钦定了A（苹果）)
ll res=comb.C(i-1,a-1);
res*=comb.C(N-i,c);
res%=mod;
ans+=res;
}
ans%=mod;
if(ans<0) ans+=mod;
cout<<ans<<"\n";
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
// cin>>T;
// while(T--){
// solve();
// }
solve();
return 0;
}
/*
AC
https://atcoder.jp/contests/abc405/submissions/66027443
*/
```
:::
## 心路历程（WA，TLE，MLE……）