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 ;
}
/********************

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

A

#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 ;
}
/********************

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

B

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 ;
}
/********************

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

C

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 ;
}
/********************

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

G

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 ;
}
/********************

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

E