Codeforces ECR52 div2翻车记

　　A：签到。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define ll long long
int t,a,b,c,d;
int main()
{
    t=read();
    while (t--)
    {
        a=read(),b=read(),c=read(),d=read();
        cout<<a/d+1ll*a/d/b*c<<endl;
    }
    return ;
}

　　B：long long。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 100010
long long n,m;
int main()
{
    cin>>n>>m;
    cout<<max(n-m*,0ll)<<' ';
    for (long long i=;i<=n;i++) if (i*(i-)/>=m) {cout<<n-i;break;}
    return ;
}

　　C：从大到小排序后贪心地修改。我猜我是唯一一个C花掉的时间比D多的。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 200010
int n,m,a[N],ans=;
int main()
{
    n=read(),m=read();
    for (int i=;i<=n;i++) a[i]=read();
    sort(a+,a+n+);reverse(a+,a+n+);
    for (int i=;a[i]!=a[n];)
    {
        int t=i;long long tot=;
        while (t<n&&tot+1ll*(a[t]-a[t+])*t<=m) tot+=(a[t]-a[t+])*t,t++;
        ans++;
        i=t;a[i]-=(m-tot)/t;
    }
    cout<<ans;
    return ;
}

　　D：做法非常显然，每个位置拆点跑个最短路然后大力dp就行了，码农题。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 12
int n,f[N*N][],g[N*N][],dis[N*N*][N*N*],dis2[N*N*][N*N*];
struct data{int x,y;
}a[N*N];
int wx[]={,,-,-,,,-,-},wy[]={,-,,-,,-,,-};
int trans(int p,int x,int y){return p*n*n+(x-)*n+y;}
void getdis()
{
    memset(dis,,sizeof(dis));memset(dis2,,sizeof(dis2));
    for (int i=;i<=n;i++)
        for (int j=;j<=n;j++)
            for (int x=;x<=n;x++)
                for (int y=;y<=n;y++)
                if (i==x||j==y) dis[trans(,i,j)][trans(,x,y)]=,dis2[trans(,i,j)][trans(,x,y)]=;
    for (int i=;i<=n;i++)
        for (int j=;j<=n;j++)
            for (int k=;k<;k++)
            if (i+wx[k]>=&&j+wy[k]>=&&i+wx[k]<=n&&j+wy[k]<=n)
            dis[trans(,i,j)][trans(,i+wx[k],j+wy[k])]=,dis2[trans(,i,j)][trans(,i+wx[k],j+wy[k])]=;
    for (int i=;i<=n;i++)
        for (int j=;j<=n;j++)
            for (int x=;x<=n;x++)
                for (int y=;y<=n;y++)
                if (i+j==x+y||i-j==x-y) dis[trans(,i,j)][trans(,x,y)]=,dis2[trans(,i,j)][trans(,x,y)]=;
    for (int i=;i<=n;i++)
        for (int j=;j<=n;j++)
            for (int x=;x<;x++)
                for (int y=;y<;y++)
                dis[trans(x,i,j)][trans(y,i,j)]=dis2[trans(x,i,j)][trans(y,i,j)]=(x!=y);
    for (int k=;k<=*n*n;k++)
        for (int i=;i<=*n*n;i++)
            for (int j=;j<=*n*n;j++)
            if (dis[i][k]+dis[k][j]<dis[i][j]||dis[i][k]+dis[k][j]==dis[i][j]&&dis2[i][k]+dis2[k][j]<dis2[i][j])
            dis[i][j]=dis[i][k]+dis[k][j],dis2[i][j]=dis2[i][k]+dis2[k][j];
}
void update(int i,int p)
{
    for (int j=;j<;j++)
    if (f[i-][j]+dis[trans(j,a[i-].x,a[i-].y)][trans(p,a[i].x,a[i].y)]<f[i][p]||
        f[i-][j]+dis[trans(j,a[i-].x,a[i-].y)][trans(p,a[i].x,a[i].y)]==f[i][p]
        &&g[i-][j]+dis2[trans(j,a[i-].x,a[i-].y)][trans(p,a[i].x,a[i].y)]<g[i][p])
        f[i][p]=f[i-][j]+dis[trans(j,a[i-].x,a[i-].y)][trans(p,a[i].x,a[i].y)],
        g[i][p]=g[i-][j]+dis2[trans(j,a[i-].x,a[i-].y)][trans(p,a[i].x,a[i].y)];
}
int main()
{
    n=read();
    for (int i=;i<=n;i++)
        for (int j=;j<=n;j++)
        {
            int x=read();
            a[x].x=i,a[x].y=j;
        }
    getdis();
    memset(f,,sizeof(f));memset(g,,sizeof(g));
    f[][]=f[][]=f[][]=g[][]=g[][]=g[][]=;
    for (int i=;i<=n*n;i++)
    update(i,),update(i,),update(i,);
    int x=;
    if (f[n*n][]<f[n*n][]||f[n*n][]==f[n*n][]&&g[n*n][]<g[n*n][]) x=;
    if (f[n*n][]<f[n*n][x]||f[n*n][]==f[n*n][x]&&g[n*n][]<g[n*n][x]) x=;
    cout<<f[n*n][x]<<' '<<g[n*n][x];
    return ;
}

　　然后就并没有时间看题了。不过E是置换并没有真的学过那大概也不算翻车了。

　　upd：于是就发现F是个傻逼题。将所有边连上缩个点dp一发即可。

　　result：rank 217 rating +7 迷之稳定。