nowcoder15550 箱庭的股市链接题解代码

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我是靠谱客的博主现代老鼠，这篇文章主要介绍nowcoder15550 箱庭的股市链接题解代码，现在分享给大家，希望可以做个参考。

链接

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题解

他给的那个递推式其实就相当于 $f_{ij}=f_{i-1,j-1}+f_{i-1,j}$

根据题意写出

$sum_{j=max(0,y-x+1)}^y binom{x-1}{y-j}$

直接暴力做居然就过了…

标签有人写了“打表题”，不是很懂…懂的大佬请在下面留言

代码

复制代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
#define mod 1'000'000'007ll
ll fact[maxn], inv[maxn], _fact[maxn];
ll C(ll n, ll m)
{
    if(n<0 or m<0 or m>n)return 0;
    return fact[n]*_fact[m]%mod*_fact[n-m]%mod;
}
int main()
{
    ll i, T, m, x, y, p, j;
    inv[1]=1;
    for(i=2;i<maxn;i++)inv[i]=inv[mod%i]*(mod-mod/i)%mod;
    fact[0]=_fact[0]=1;
    rep(i,maxn-1)fact[i]=fact[i-1]*i%mod, _fact[i]=_fact[i-1]*inv[i]%mod;
    while(~scanf("%lld%lld%lld%lld",&m,&x,&y,&p))
    {
        ll ans=0;
        for(j=max(0ll,y-x+1);j<=y;j++)(ans+=C(x-1,y-j))%=mod;
        ans=ans*p%mod;
        printf("%lldn",ans);
    }
    return 0;
}

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#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
#define mod 1'000'000'007ll
ll fact[maxn], inv[maxn], _fact[maxn];
ll C(ll n, ll m)
{
    if(n<0 or m<0 or m>n)return 0;
    return fact[n]*_fact[m]%mod*_fact[n-m]%mod;
}
int main()
{
    ll i, T, m, x, y, p, j;
    inv[1]=1;
    for(i=2;i<maxn;i++)inv[i]=inv[mod%i]*(mod-mod/i)%mod;
    fact[0]=_fact[0]=1;
    rep(i,maxn-1)fact[i]=fact[i-1]*i%mod, _fact[i]=_fact[i-1]*inv[i]%mod;
    while(~scanf("%lld%lld%lld%lld",&m,&x,&y,&p))
    {
        ll ans=0;
        for(j=max(0ll,y-x+1);j<=y;j++)(ans+=C(x-1,y-j))%=mod;
        ans=ans*p%mod;
        printf("%lldn",ans);
    }
    return 0;
}