2018.10.20 XMYZ Day1总结

上周的忘写了……题目没有作者……

T1.backpack

期望得分100，实际得分100.

感觉我自己真是不如以前了……以前做这种题都是秒掉的，现在怎么想了10分钟啊……

因为物品的体积和价值都非常小，我们有一句套话，“远距离贪心，近距离暴力”，所以虽然背包的体积特别大，我们可以把他压缩成1000000左右，剩下的直接暴力取性价比最高的即可。

看一下代码。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')

using namespace std;
typedef long long ll;
const int M = ;
const int INF = ;
const ll mod = 1e9+;

ll read()
{
    ll ans = ,op = ;
    char ch = getchar();
    while(ch < '' || ch > '')
    {
        if(ch == '-') op = -;
        ch = getchar();
    }
    while(ch >= '' && ch <= '')
    {
        ans *= ;
        ans += ch - '';
        ch = getchar();
    }
    return ans * op;
}

ll dp[M],m,n,a[M],b[M],v[],mpos,ans;
double maxn;

void solve1()
{
    rep(i,,n)
    rep(j,a[i],m) dp[j] = max(dp[j],dp[j-a[i]] + b[i]);
    printf("%lld\n",dp[m]);
}

int main()
{
    freopen("backpack.in","r",stdin);
    freopen("backpack.out","w",stdout);
    n = read(),m = read();
    rep(i,,n)
    {
        a[i] = read(),b[i] = read();
        v[a[i]] = max(b[i],v[a[i]]);
    }
    if(n <=  && m <= ) solve1();
    else
    {
        rep(i,,)
        {
            double p = (double)(i),q = (double)(v[i]);
            if(q / p > maxn) maxn = q / p,mpos = i;
        }
        if(m > )
        {
            ll k = (m - ) / mpos;
            m -= k * mpos,ans += k * v[mpos];
        }
        rep(i,,)
        rep(j,i,m) dp[j] = max(dp[j],dp[j-i] + v[i]);
        ans += dp[m];
        printf("%lld\n",ans);
    }
    return ;
}

T2.sort

期望得分20，实际得分0.

这道题考试的时候没怎么想，主要刚T3了，而且我也不会求方案数。写了个爆搜，然后GG了。

后来想到这个最大值和最小值很好维护，最大值取后一半减去前一半即可，而最小值的话只要奇偶分别维护取差值即可。这个可以使用线段树分别维护奇偶，稍微有点麻烦，下放的时候有很多特判。

之后至于方案数，mrclr大佬说，你可以把它看成一个括号序列匹配，就是一个卡特兰数。求一下阶乘和阶乘的逆元即可。

看一下代码。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')

using namespace std;
typedef long long ll;
const int M = ;
const int INF = ;
const ll mod = 1e9+;

ll read()
{
    ll ans = ,op = ;
    char ch = getchar();
    while(ch < '' || ch > '')
    {
        if(ch == '-') op = -;
        ch = getchar();
    }
    while(ch >= '' && ch <= '')
    {
        ans *= ;
        ans += ch - '';
        ch = getchar();
    }
    return ans * op;
}

struct seg
{
    ll v,ov,ev,lazy;
}t[M<<];

ll n,m,c[M<<],l,r,op,val,sum[M<<],inv[M<<];

ll qpow(ll a,ll b)
{
    ll p = ;
    while(b)
    {
        if(b&1LL) p *= a,p %= mod;
        a *= a,a %= mod;
        b >>= 1LL;
    }
    return p;
}

void init()
{
    sum[] = inv[] = inv[] = 1ll;
    rep(i,,n<<) sum[i] = sum[i-] * i,sum[i] %= mod;
    inv[n<<] = qpow(sum[n<<],mod-);
    per(i,(n<<)-,) inv[i] = inv[i+] * (i+),inv[i] %= mod;
    //rep(i,1,n<<1) printf("%lld ",sum[i]);enter;
}

void pushup(int p)
{
    t[p].v = t[p<<].v + t[p<<|].v,t[p].v %= mod;
    t[p].ev = t[p<<].ev + t[p<<|].ev,t[p].ev %= mod;
    t[p].ov = t[p<<].ov + t[p<<|].ov,t[p].ov %= mod;
}

void pushdown(int p,int l,int r)
{
    ll mid = (l+r) >> ,k = mid - l + ,q = r - mid;
    t[p<<].v += t[p].lazy * k,t[p<<].v %= mod;
    t[p<<|].v += t[p].lazy * q,t[p<<|].v %= mod;
    t[p<<].lazy += t[p].lazy,t[p<<|].lazy += t[p].lazy;
    t[p<<].lazy %= mod,t[p<<|].lazy %= mod;
    if(k & )
    {
        if(l&) t[p<<].ov += t[p].lazy * (k +  >> ),t[p<<].ev += t[p].lazy * (k>>);
        else t[p<<].ov += t[p].lazy * (k >> ),t[p<<].ev += t[p].lazy * (k +  >> );
    }
    else t[p<<].ov += t[p].lazy * (k >> ),t[p<<].ev += t[p].lazy * (k >> );
    t[p<<].ov %= mod,t[p<<].ev %= mod;
    if(q & )
    {
        if((mid + ) & ) t[p<<|].ov += t[p].lazy * (q +  >> ),t[p<<|].ev += t[p].lazy * (q >> );
        else t[p<<|].ov += t[p].lazy * (q >> ),t[p<<|].ev += t[p].lazy * (q +  >> );
    }
    else t[p<<|].ov += t[p].lazy * (q >> ),t[p<<|].ev += t[p].lazy * (q >> );
    t[p<<|].ov %= mod,t[p<<|].ev %= mod;
    t[p].lazy = ;
}

void build(int p,int l,int r)
{
    if(l == r)
    {
        t[p].v = c[l];
        if(l & ) t[p].ov = c[l];
        else t[p].ev = c[l];
        return;
    }
    int mid = (l+r) >> ;
    build(p<<,l,mid),build(p<<|,mid+,r);
    pushup(p);
}

void modify(int p,int l,int r,int kl,int kr,ll val)
{
    if(l == kl && r == kr)
    {
        ll k = r - l + ;
        t[p].v += val * k,t[p].lazy += val;
        t[p].v %= mod,t[p].lazy %= mod;
        if(k & )
        {
            if(l & ) t[p].ov += val * (k +  >> ),t[p].ev += val * (k >> );
            else t[p].ov += val * (k >> ),t[p].ev += val * (k +  >> );
        }
        else t[p].ov += val * (k >> ),t[p].ev += val * (k >> );
        t[p].ov %= mod,t[p].ev %= mod;
        return;
    }
    if(t[p].lazy) pushdown(p,l,r);
    int mid = (l+r) >> ;
    if(kr <= mid) modify(p<<,l,mid,kl,kr,val);
    else if(kl > mid) modify(p<<|,mid+,r,kl,kr,val);
    else modify(p<<,l,mid,kl,mid,val),modify(p<<|,mid+,r,mid+,kr,val);
    pushup(p);
}

ll query(int p,int l,int r,int kl,int kr,int f)
{
    if(l == kl && r == kr)
    {
        if(f == ) return t[p].v;
        else if(f == ) return t[p].ov;
        else if(f == ) return t[p].ev;
    }
    ll mid = (l+r) >> ;
    if(t[p].lazy) pushdown(p,l,r);
    if(kr <= mid) return query(p<<,l,mid,kl,kr,f);
    else if(kl > mid) return query(p<<|,mid+,r,kl,kr,f);
    else return query(p<<,l,mid,kl,mid,f) + query(p<<|,mid+,r,mid+,kr,f) % mod;
}

int main()
{
    freopen("sort.in","r",stdin);
    freopen("sort.out","w",stdout);
    n = read(),m = read();
    init();
    rep(i,,n<<) c[i] = read();
    build(,,n<<);
    rep(i,,m)
    {
        op = read();
        if(op == ) l = read(),r = read(),val = read(),modify(,,n<<,l,r,val);
        else
        {
            l = read(),r = read();
            int g = (l + r) >> ,h = (r - l + ) >> ;
            ll a = query(,,n<<,l,g,),b = query(,,n<<,g+,r,),maxx = (b - a + mod << ) % mod;
            ll ka = query(,,n<<,l,r,),kb = query(,,n<<,l,r,),minx = abs(kb - ka) % mod;
            ll cat = sum[h << ] * inv[h] % mod * inv[h+] % mod;
            printf("%lld %lld %lld\n",maxx >> ,minx,cat);
        }
    }
    return ;
}

T3.digit

期望得分20，实际得分20.

考试的时候主要刚这题了……想到使用dp[i][j]表示前i位对4，5，6取模后结果为j的情况数，发现这个对于每一位都是一样的，可以使用矩乘优化计算。但是我的DP一直是错的……到死也没de出来。

后来发现我的计算是不对的，不应该对4，5，6分别计算，而应该对它们的lcm60取模，这样的话我们使用矩乘优化，之后取所有能整除4，5，6的数的结果即可。

看一下代码。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')

using namespace std;
typedef long long ll;
const int M = ;
const int INF = ;
const ll mod = 1e9+;

ll read()
{
    ll ans = ,op = ;
    char ch = getchar();
    while(ch < '' || ch > '')
    {
    if(ch == '-') op = -;
    ch = getchar();
    }
    while(ch >= '' && ch <= '')
    {
    ans *= ;
    ans += ch - '';
    ch = getchar();
    }
    return ans * op;
}

struct matrix
{
    ll a[][];
    void init(int x)
    {
    memset(a,,sizeof(a));
    rep(i,,M-) a[i][i] = x;
    }
    matrix operator * (const matrix &g) const
    {
    matrix p;
    p.init();
    rep(i,,M-)
    rep(j,,M-)
    rep(k,,M-) p.a[i][j] += a[i][k] * g.a[k][j],p.a[i][j] %= mod;
    return p;
    }
    matrix operator *= (const matrix &g)
    {
    return *this = *this * g;
    }
    matrix operator ^ (ll g)
    {
    matrix p,q = *this;
    p.init();
    while(g)
    {
        if(g & ) p *= q;
        q *= q;
        g >>= ;
    }
    return p;
    }
}f,g;

ll l,r,k,ans;

int main()
{
    l = read(),r = read(),k = read();
    f.init();
    rep(i,,M-)
    {
    rep(j,,M-) f.a[j][i] = k / M % mod;
    rep(j,,k%M-) f.a[(i+j) % M][i]++,f.a[(i+j) % M][i] %= mod;
    }
    g = f ^ r;
    rep(i,,M-) if(i %  ==  || i %  ==  || i %  == ) ans += g.a[i][],ans %= mod;
    g = f ^ (l - );
    rep(i,,M-) if(i %  ==  || i %  ==  || i %  == ) ans += (mod - g.a[i][]),ans %= mod;
    printf("%lld\n",ans);
    return ;
}