湖南省第十七届大学生计算机程序设计竞赛（HNCPC2021

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A.签到题，用map嵌个结构体。

B.对于满足与即可

C.直接暴力做，但用map被卡常，而数据又有点大,考虑到题面保证给的数据是递增的，用二分查找来判断是否存在。

D.先处理好的逆元env，然后对于每个问询，先算出，然后每次乘于左端逆元与新右端点。需注意

A: 
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struct  A
{
    int c,j,z;
};

LL t;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>t;
    f(sb,1,t){
        map<char ,A> m;
        f(i,1,15){
            char na,nb;
            int a,b,c,d;
            cin>>na>>nb>>a>>b>>c>>d;
            if(a>b){
                m[na].c+=3,m[na].j+=(a-b),m[na].z+=a;
                m[nb].j+=(b-a),m[nb].z+=b;
            }
            else if(a==b){
                m[na].c+=1,m[na].j+=(a-b),m[na].z+=a;
                m[nb].c+=1,m[nb].j+=(b-a),m[nb].z+=b;
            }
            else{
                 m[nb].c+=3,m[na].j+=(a-b),m[na].z+=a;
                 m[nb].j+=(b-a),m[nb].z+=b;
            }
            a=c,b=d;
            if(a>b){
                m[na].c+=3,m[na].j+=(a-b),m[na].z+=a;
                m[nb].j+=(b-a),m[nb].z+=b;
            }
            else if(a==b){
                m[na].c+=1,m[na].j+=(a-b),m[na].z+=a;
                m[nb].c+=1,m[nb].j+=(b-a),m[nb].z+=b;
            }
            else{
                 m[nb].c+=3,m[na].j+=(a-b),m[na].z+=a;
                m[nb].j+=(b-a),m[nb].z+=b;
            }
        }
        f(i,0,5){
            char d='A'+i;
            cout<<m[d].c<<" "<<m[d].j<<" "<<m[d].z<<"\n";
        }
    }
    return 0;
}

B:

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LL tt, n, m, r, c, v;
int main()
{
    while (scanf("%lld%lld%lld%lld%lld", &n, &m, &r, &c, &v) == 5)
    {
        if (r * c <= v && n * m - v + 1 >= (n - r + 1) * (m - c + 1))
        {
            cout << "Yes\n";
        }
        else
            cout << "No\n";
    }
    return 0;
}

C:

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LL n, x;
int main()
{
    while (scanf("%lld", &n) == 1)
    {
        vi tar, a;
        f(i, 0, n - 1) scanf("%lld", &x), tar.pb(2 * x), a.pb(x);
        for (auto s = tar.begin(); s != tar.end();)
        {
            int flag = 1;
            f(i, 0, n-1)
            {
                if (2 * a[i] != *s)
                {
                    auto dd = lower_bound(all(a), *s - a[i]);
                    if (dd != a.end() && *dd == *s - a[i])
                    {
                        s = tar.erase(s);
                        flag = 0;
                        break;
                    }
                }
            }
            if (flag)
                s += 1;
        }
        cout<<n-tar.size() << "\n";
    }
    return 0;
}

D:

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const LL pp = 1e9 + 7,N=1e6+9;
LL n, m,env[N];

LL pow_mod(LL a, LL b, LL p)
{
    LL ret = 1;
    while (b)
    {
        if (b & 1)
            ret = (ret * a) % p;
        a = (a * a) % p;
        b >>= 1;
    }
    return ret;
}
LL Fermat(LL a, LL p)
{ 
    return pow_mod(a, p - 2, p);
}

int main()
{
    f(i,1,1e6){
        env[i]=Fermat(i,pp);
    }
    while (scanf("%lld%lld", &n, &m) == 2)
    {
        LL ans=0,te;
        te = 1;
        f(i, 1, m)
        {
            te *= i;
            te %= pp;
        }
        ans += te;
        ans %= pp;
        f(j, 1, n - 1)
        {

            te *= env[j];
            te %= pp;
            te *= (j + m);
            te %= pp;
            ans += te;
            ans %= pp;
        }
        cout << ans << "\n";
    }
    return 0;
}