2016ACM/ICPC亚洲区沈阳站

emm，a出3题，补了两题

A，B水题

#include<bits/stdc++.h>

#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define read(a) scanf("%d",&a)

#define rread(a,b) scanf("%d%d",&a,&b)

#define pii pair<int,int>

#define C 0.5772156649

#define pi acos(-1.0)

#define ll long long

#define mod 1000000007

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f3f;

int main()

{

    int n;

    read(n);

    for(int i=;i<n;i++)

    {

        int a,b;

        rread(a,b);

        printf("%d\n",*max(a,b)+min(a,b));

    }

    return ;

}

/********************

********************/

#include<bits/stdc++.h>

#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define read(a) scanf("%d",&a)

#define rread(a,b) scanf("%d%d",&a,&b)

#define pii pair<int,int>

#define C 0.5772156649

#define pi acos(-1.0)

#define ll long long

#define mod 1000000007

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f3f;

char s[N];

int main()

{

    int n;

    read(n);

    for(int i=;i<n;i++)

    {

        int a,b;

        scanf("%s",&s);

        int len=strlen(s),ans=;

        for(int i=;i<len;i++)

        {

            if(s[i]=='H')ans++;

            else if(s[i]=='C')ans+=;

            else ans+=;

        }

        printf("%d\n",ans);

    }

    return ;

}

/********************

********************/

C：递推，首先想到矩阵快速幂，构造一个7维的矩阵，分别是ai-1，ai-2，i^4,i^3,i^2,i,1,再转移下去

#include<bits/stdc++.h>

#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define read(a) scanf("%d",&a)

#define rread(a,b) scanf("%d%d",&a,&b)

#define pii pair<int,int>

#define C 0.5772156649

#define pi acos(-1.0)

#define ll long long

#define mod 2147493647

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f3f;

struct Node{

    int len;

    ll a[N][N];

};

Node mul(Node x,Node y)

{

    Node ans;

    memset(ans.a,,sizeof ans.a);

    ans.len=x.len;

    for(int i=;i<x.len;i++)

    {

        for(int j=;j<x.len;j++)

        {

            for(int k=;k<y.len;k++)

            {

                ans.a[i][k]=(ans.a[i][k]+x.a[i][j]*y.a[j][k]%mod)%mod;

            }

        }

    }

    return ans;

}

Node quick_mul(Node x,ll n)

{

    Node ans;

    memset(ans.a,,sizeof ans.a);

    ans.len=x.len;

    for(int i=;i<x.len;i++)ans.a[i][i]=;

    while(n)

    {

        if(n&)ans=mul(ans,x);

        x=mul(x,x);

        n/=;

    }

    return ans;

}

int main()

{

    int t;

    read(t);

    for(int i=;i<t;i++)

    {

        ll n,x,y;

        scanf("%lld%lld%lld",&n,&x,&y);

        if(n==)

        {

            printf("%lld\n",x%mod);

            continue;

        }

        if(n==)

        {

            printf("%lld\n",y%mod);

            continue;

        }

        Node A;

        A.len=;

        A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;

        A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;

        A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;

        A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;

        A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;

        A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;

        A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;

        A=quick_mul(A,n-);

        ll ans=A.a[][]*y%mod;

        ans=(ans+A.a[][]*x%mod)%mod;

        ans=(ans+A.a[][]****%mod)%mod;

        ans=(ans+A.a[][]***%mod)%mod;

        ans=(ans+A.a[][]**%mod)%mod;

        ans=(ans+A.a[][]*%mod)%mod;

        ans=(ans+A.a[][]%mod)%mod;

        printf("%lld\n",ans);

    }

    return ;

}

/********************

********************/

G是个二分+积分题，由于关系找错了，导致一直没过，二分水平面和竖直的夹角，用积分求体积和，不倾斜的体积比较

#include<bits/stdc++.h>

#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define read(a) scanf("%d",&a)

#define rread(a,b) scanf("%d%d",&a,&b)

#define pii pair<int,int>

#define C 0.5772156649

#define pi acos(-1.0)

#define ll long long

#define mod 2147493647

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f3f;

double f(double x)

{

    double y=acos(*tan(x)-);

    double a1,a2;

    a1=pi*cos(y)-y*cos(y)+sin(y)-(1.0/3.0)*sin(y)*sin(y)*sin(y);

    y=pi;

    a2=pi*cos(y)-y*cos(y)+sin(y)-(1.0/3.0)*sin(y)*sin(y)*sin(y);

    return (1.0/tan(x))*(a1-a2);

}

double bs(double x)

{

    double l=,r=pi/;

    while(r-l>1e-)

    {

        double m=(l+r)/;

        if(f(m)>x)r=m;

        else l=m;

    }

    return l;

}

int main()

{

    int t;

    scanf("%d",&t);

    for(int i=;i<t;i++)

    {

        double d;

        scanf("%lf",&d);

        if(d<1e-)

        {

            printf("%.5f\n",0.0);

            continue;

        }

        double ans=pi*sqrt(+(-d)*(-d));

        if(d<)

        {

            double ang=bs(d*pi);

            double p=*tan(ang),ang1=acos(-p),v=ang1-(-p)*sqrt(-(-p)*(-p));

            ans=v/sin(ang);

        }

        printf("%.5f\n",ans);

    }

    return ;

}

/********************

********************/

E，求点为m个的团有多少个，因为没学过最大团的算法，所以不敢写，补题的时候发现居然dfs就能过= =，直接暴力dfs找到含m个点的团就退出

#include<bits/stdc++.h>

#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define read(a) scanf("%d",&a)

#define rread(a,b) scanf("%d%d",&a,&b)

#define pii pair<int,int>

#define C 0.5772156649

#define pi acos(-1.0)

#define ll long long

#define mod 2147493647

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f3f;

bool ok[N][N];

int contain[N],ans;

int n,m,s;

vector<int>v[N];

void dfs(int u,int a[],int sz)

{

    if(sz==s)

    {

        ans++;

        return ;

    }

    for(int i=;i<v[u].size();i++)

    {

        bool f=;

        for(int j=;j<sz;j++)

        {

            if(!ok[v[u][i]][a[j]])

            {

                f=;

                break;

            }

        }

        if(!f)

        {

            a[sz++]=v[u][i];

            dfs(v[u][i],a,sz);

            sz--;

        }

    }

}

int main()

{

    int t;

    read(t);

    for(int i=;i<t;i++)

    {

        rread(n,m);read(s);

        for(int i=;i<=n;i++)v[i].clear();

        memset(ok,,sizeof ok);

        while(m--)

        {

            int a,b;

            rread(a,b);

            if(a>b)swap(a,b);

            v[a].pb(b);

            ok[a][b]=ok[b][a]=;

        }

        ans=;

        for(int i=;i<=n;i++)

        {

           // memset(contain,0,sizeof contain);

            contain[]=i;

            dfs(i,contain,);

        }

        printf("%d\n",ans);

    }

    return ;

}

/********************

********************/

巴特西

2016ACM/ICPC亚洲区沈阳站

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