Codeforces round #254 (Div. 1)-a, B

A: select the 2.1 side. Obviously, you can figure it out...

But at first I read the wrong question: Sad ,,,,

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>#include<math.h>using namespace std;#define eps 1e-6#define zero(x) ((fabs(x)<eps?0:x))#define maxn 550int val[maxn];int main(){    int n,m,u,v,w;    while(~scanf("%d%d",&n,&m))    {        for(int i=1;i<=n;i++)        {            scanf("%d",&val[i]);        }        double ans=0.0;        for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&w);            double pp=0;            pp=1.0*(val[u]+val[v])/(w);            if(pp>ans)ans=pp;        }        printf("%.10lf\n",ans);    }    return 0;}

B:

Suppose d <= SQRT (n ):

Then we use each 1 in array B to match array;

Complexity N * SQRT (n ).

Otherwise:

We use numbers to match each location to see if the person in this location exists.

Theoretically feasible complexity

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>#include<math.h>using namespace std;#define eps 1e-6#define zero(x) ((fabs(x)<eps?0:x))#define maxn 110000#define LL long longint val[maxn];int n,d;LL x;int a[maxn];int b[maxn];int getNextX() {    x = (x * 37 + 10007) % 1000000007;    return x;}int initAB() {    int i;    for(i = 0; i < n; i = i + 1){        a[i] = i + 1;    }    for(i = 0; i < n; i = i + 1){        swap(a[i], a[getNextX() % (i + 1)]);    }    for(i = 0; i < n; i = i + 1){        if (i < d)            b[i] = 1;        else            b[i] = 0;    }    for(i = 0; i < n; i = i + 1){        swap(b[i], b[getNextX() % (i + 1)]);    }}int c[maxn];int ip[maxn];struct list{    int id;    int next;    int pre;}node[maxn];vector<int>vec;int main(){    while(~scanf("%d%d%lld",&n,&d,&x))    {        vec.clear();        initAB();        for(int i=0;i<=n+1;i++)        {            node[i].id=i;            node[i].next=i+1;            node[i].pre=i-1;        }        for(int i=n;i>=1;i--)        {            a[i]=a[i-1];            b[i]=b[i-1];        }        int m=sqrt(n);        for(int i=1;i<=n;i++)        {            ip[a[i]]=i;        }        if(m<=d)        {            for(int i=n;i>=1;i--)            {                int z=ip[i];                for(int j=0;j!=n+1;j=node[j].next)                {                    int y=node[j].id;                    if(y<z)continue;                    if(b[y-z+1])                    {                        c[y]=i;                        node[node[j].next].pre=node[j].pre;                        node[node[j].pre].next=node[j].next;                    }                }            }        }        else        {            for(int i=1;i<=n;i++)            {                if(b[i])                {                    vec.push_back(i);                }            }            for(int i=0;i<vec.size();i++)            {                int y=vec[i];                for(int j=1;j<=n-y+1;j++)                {                    c[j+y-1]=max(c[j+y-1],a[j]);                }            }        }        for(int i=1;i<=n;i++)        {            printf("%d\n",c[i]);        }    }    return 0;}

Codeforces round #254 (Div. 1)-a, B