AtCoder Regular Contest 107(VP)

比赛体验良好，网站全程没有挂。题面简洁好评，题目质量好评。对于我这个蒟蒻来说非常合适的一套题目。

A. Simple Math

Description

Given are three positive integers $A,B,$ and $C$ . Compute the following value modulo $998244353:\sum_{a=1}^A\sum_{b=1}^B\sum_{c=1}^C a\times b\times c.$

$1\leq A,B,C\leq 1e9$

Solution

\[\sum_{a=1}^A\sum_{b=1}^B\sum_{c=1}^C a\times b\times c
= \sum_{a=1}^A a\sum_{b=1}^B b\sum_{c=1}^C c
\]

疯狂取模就好了。

Code

赛时用时 5min。

//Author: RingweEH
const ll mod=998244353;
int main()
{
    ll A=read(),B=read(),C=read();
    A%=mod; B%=mod; C%=mod;
    A=A*(A+1)/2%mod; B=B*(B+1)/2%mod; C=(C*(C+1))/2%mod;
    ll ans=A*B%mod*C%mod;
    printf( "%lld",ans );
}

B. Quadruple

Problem Link

Description

Given are integers $N$ and $K$ . How many quadruples of integers $(a,b,c,d)$ satisfy both of the following conditions?

$1\leq a,b,c,d\leq N$
$a+b-c-d=K$

$1\leq N\leq 1e5,-2(N-1)\leq K\leq 2(N-1)$

Solution

暴力枚举显然非常的不可行。按照正负，把原来的式子分成两部分：

$a+b=x$
$c+d=x-K$

然后对于每一部分，设 $f(N,K)$ 表示 $1\leq a,b\leq N$ 时 $a+b=K$ 的方案数。于是得到：

\[f(N,K)=\min(K-1,2\times N+1-K)
\]

然后 $O(N)$ 算一下就好了。不过容斥可以做到 $O(1)$ 。

注意特判 $f(N,K)$ 中 $K$ 不合法的情况。

Code

赛时用时 40min.

//Author:RingweEH
ll n,k;
ll calc( ll n,ll k )
{
    if ( k<2 ) return 0;
    if ( k>2*n ) return 0;
    return min( k-1,2*n+1-k );
}
int main()
{
    n=read(); k=read();
    ll ans=0;
    for ( int i=2; i<=2*n; i++ )
        ans=ans+calc( n,i )*calc( n,i-k );
    printf( "%lld",ans );
}

C. Shuffle Permutation

Problem Link

Description

Given are an $N\times N$ matrix and an integer $K$. The entry in the $i$-th row and $j$-th column of this matrix is denoted as $a_{i,j}$ . This matrix contains each of $1,2,...,N^2$ exactly once.

Sigma can repeat the following two kinds of operation arbitrarily many times in any order.

Pick two integers $x,y(1\leq x<y\leq N)$ that satisfy $a_{i,x}+a_{i,y}\leq K$ for all $i(1\leq i\leq N)$ and swap the $x$-th and the $y$-th columns.
Pick two integers $x,y(1\leq x<y\leq N)$ that satisfy $a_{x,i}+a_{y,i}\leq K$ for all $i(1\leq i\leq N)$ and swap the $x$-th and the $y$-th rows.

How many matrices can he obtain by these operations?Find it modulo $998244353.$

$1\leq N\leq 50. 1\leq K\leq 2\times N^2.$

Solution

啥都不知道，开始手模样例。

3min之后……发现自己是个思博。显然行列之间没什么关系。所以乘起来就好了。

然后呢？还是不知道行列怎么算。

仔细思考时候发现，如果把行（或者列）当成点，能够 swap 就连边，会发现凡是联通块内点都能任意互换。

A-B-C => B-A-C => B-C-A => C-B-A

以上是列的 $N=3$ 情况。那么只需要暴力枚举判断加边，带权并查集就好了（事实上直接统计也行）。

Code

赛时用时 40min.

WA 了一发，原因是把计算行/列的情况中乘法原理搞成了加法……就这还能 AC $\dfrac{22}{25}$ ……没话讲。

//Author:RingweEH
//由于代码具有强对称性，删减了部分 copy 的函数使得代码看上去更加简短（
int find_r( int x )
{
    return x==fa_r[x] ? x : fa_r[x]=find_r( fa_r[x] );
}
void merge_r( int x,int y )
{
    int fx=find_r(x),fy=find_r(y);
    if ( fx==fy ) return;
    fa_r[fx]=fy;
}
void init()
{
    fac[0]=1;
    for ( int i=1; i<=50; i++ )
        fac[i]=fac[i-1]*i%mod;
}
int main()
{
    for ( int i=1; i<=n; i++ )
        fa_c[i]=i,fa_r[i]=i;
    for ( int i=1; i<=n; i++ )
     for ( int j=i+1; j<=n; j++ )
    {
        bool fl=1;
        for ( int k=1; k<=n; k++ )
            if ( a[k][i]+a[k][j]>K ) fl=0;
        if ( fl ) merge_r( i,j );
    }
    memset( cntr,0,sizeof(cntr) ); memset( cntc,0,sizeof(cntc) );
    for ( int i=1; i<=n; i++ )
        cntr[find_r(i)]++,cntc[find_c(i)]++;
    ll ans1=1,ans2=1; init();
    for ( int i=1; i<=n; i++ )
    {
        if ( cntr[i]>1 ) ans1=(ans1*fac[cntr[i]])%mod;
        if ( cntc[i]>1 ) ans2=(ans2*fac[cntc[i]])%mod;
    }
    printf( "%lld\n",ans1*ans2%mod );
}

D.Number of Multisets

Problem Link

Description

You are given two positive integers $N$ and $K$ . How many multisets of raional numbers satisfy all of the following conditions?

The multiset has exactly $N$ elements and the sum of them is equal to $K$.
Each element of the multiset is one of $1,\dfrac{1}{2},\dfrac{1}{4},...$ . In other words,each element can be represented as $\dfrac{1}{2^i}(i=0,1,...)$

The answer may be large,so print it modulo $998244353.$

Solution

小清新分讨DP题。

分为两类：用 $1$ 的和不用的。

对于用了至少一个 $1$ 的，$f[i][j]=f[i-1][j-1]$

对于没有用 $1$ 的，$f[i][j]=f[i][2\times j]$

显然把所有元素乘2就是等价的了。

综上， $f[i][j]=f[i-1][j-1]+f[i][2\times j]$.

最后考虑边界就好了： $f[i][j]=0(i<j)$

时间复杂度 $O(N^2)$

Code

//Author: RingweEH
int main()
{
    n=read(); k=read();
    f[0][0]=1;
    for ( int i=1; i<=n; i++ )
     for ( int j=i; j; j-- )
        f[i][j]=(f[i-1][j-1]+(j*2>i ? 0 : f[i][j*2]) )%mod;
    printf( "%lld",f[n][k] );
}

E. Mex Mat

Problem Link

Description

Consideer an $N\times N$ matrix, Let us denote by $a_{i,j}$ the entry in the $i$-th row and $j$-th column. For $a_{i,j}$ where $i=1$ or $j=1$ holds,its value is one of $0,1,2$ and given in the input. The remaining entries are defined as follows:

$a_{i,j}=mex(a_{i-1,j},a_{i,j-1})(2\leq i,j\leq N)$

( $mex$ 定义见题面)

How many entries of the matrix are $0,1,2$ respectively ?

Solution

一开始看成了SG函数的那个 mex emmm

神仙结论题……官方题解简洁而优雅暴力。

官方题解给了一个 $20\times 20$ 的随机矩阵的完整形式，然后发现在 $i>4,j>4$ 之后都有 $a_{i,j}=a_{i-1,j-1}$ ，暴力跑前面的四行四列即可。

证明……你可以暴力跑 $n=5$ 的矩阵，然后有 $a_{4,4}=a_{5,5}$ ，按这个性质下去就好了。

Code

REH 这个zz又写错了一次边界……

//Author: RingweEH
        for ( int i=2; i<=10; i++ )
        {
            a[i-1]=b[i];
            for ( int j=i; j<=n; j++ )
                cnt[a[j]=mex[a[j]][a[j-1]]]++;
            b[i]=a[i];
            for ( int j=i+1; j<=n; j++ )
                cnt[b[j]=mex[b[j]][b[j-1]]]++;
        }

F. Sum of Abs

Problem Link

Description

Given is a simple undirected graph with $N$ vertices and $M$ edges. Its vertices are numbered $1,2,...,N$ and its edges are numbered $1,2,..,M$ .On Vertex $i(1\leq i\leq N)$ two integers $A_i$ and $B_i$ are written. Edge $i(1\leq i\leq M)$ connects Vertices $U_i$ and $V_i$ .

Snuke picks zero or more vertices and delete them. Deleting Vertex $i$ costs $A_i$ . When a vertex is deleted,edges that are incident to the vertex are also deleted. The score after deleting vertices is calculated as follows:

The score is the sum of the scores of all connected components.
The score of a connected component is the absolute value of the sum of $B_i$ of the vertices in the connected component.

Find the maxinum possible profit Snuke can gain. profit is (score)-(the sum of costs).

Solution

很好的网络流建模练习题。

简化题意：$n$ 点 $m$ 边的无向图，删点代价 $A_i$ ，最大化所有连通块 $B_i$ 之和的绝对值之和减去总代价。

价值就等价于给每个点一个系数 $p_i=\pm 1$ ，使得 $\forall (x,y)\in E,p_x=p_y$ 的最大的 $\sum_{i=1}^n p_ib_i$

考虑网络流建图：将 $b_i$ 分为正负两类， $S$ 向正的连 $2b_i$ 的边，负的向 $T$ 连 $-2b_i$ 的边，图中直接相连的两点连上 $\infty$ （如果两点不同向，必须有一个被割掉）

再来看删除操作。可以看做这个点不要求正负，且不帮助连边。那么进行拆点，建立一条 $(i,i+n)\in E'$ ，流量为 $a_i+|b_i|$ 表示删除的代价。

对于一条原图中的边，连 $(x+n,y),(y+n,x)$ ，此时若 $(x,x+n)$ 被删掉，那么 $x$ 就到不了 $y+n$ .

最终答案就是 $\sum_{i=1}^n |b_i|$ 减去最小割。

Code

//Author: RingweEH
void add( int u,int v,int w )
{
    e[tot].to=v; e[tot].nxt=head[u]; head[u]=tot; e[tot].val=w; tot++;
    if ( tot&1 ) add( v,u,0 );
}
bool bfs()
{
    memset( dis,inf,sizeof(dis) ); dis[0]=0; q.push( 0 );
    while ( !q.empty() )
    {
        int u=q.front(); q.pop();
        for ( int i=head[u]; i!=-1; i=e[i].nxt )
        {
            int v=e[i].to;
            if ( !e[i].val || (dis[v]!=inf) ) continue;
            dis[v]=dis[u]+1; q.push( v );
        }
    }
    return dis[2*n+1]!=inf;
}
int dfs( int u,int res )
{
    if ( u>2*n ) return res;
    for ( int i=head[u]; i!=-1; i=e[i].nxt )
    {
        int v=e[i].to;
        if ( !e[i].val || (dis[v]!=dis[u]+1) ) continue;
        int tmp=dfs( v,min(res,e[i].val) );
        if ( tmp ) { e[i].val-=tmp,e[i^1].val+=tmp; return tmp; }
    }
    return 0;
}
int Dinic()
{
    int num,res=0;
    while ( bfs() )
        while ( num=dfs( 0,inf ) ) res+=num;
    return res;
}
int main()
{
    n=read(); m=read();
    for ( int i=1; i<=n; i++ )
        a[i]=read();
    memset( head,-1,sizeof(head) ); int ans=0;
    for ( int i=1,x; i<=n; i++ )
    {
        x=read(); ans+=abs(x);
        add( i,i+n,abs(x)+a[i] );
        if ( x>=0 ) add( 0,i,2*x );
        else add( i+n,2*n+1,-2*x);
    }
    for ( int i=1,u,v; i<=m; i++ )
        u=read(),v=read(),add( u+n,v,inf ),add( v+n,u,inf );
    printf( "%d\n",ans-Dinic() );
}