2012 jinhua

2012 jinhua

I题：签到题，没什么意思。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstdlib>

using namespace std;

const int maxn=1000100;

int n;
int a[maxn];

int main()
{
    while(cin>>n&&n){
        int ans=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            ans+=a[i]*a[i];
        }
        cout<<ans<<endl;
    }
    return 0;
}

J题：排列组合水题。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>

using namespace std;

const int maxn=1100;

typedef long long ll;
ll N,M,K;
ll P;
ll L[maxn],R[maxn];
char s[30];int x;
char t[30];int y;
bool visL[maxn][maxn],visR[maxn][maxn];

int main()
{
   // freopen("in.txt","r",stdin);
    while(cin>>N>>M>>K){
        if(N==0&M==0&K==0) break;
        cin>>P;
        memset(L,0,sizeof(L));
        memset(R,0,sizeof(R));
        memset(visL,0,sizeof(visL));
        memset(visR,0,sizeof(visR));
        while(P--){
            scanf("%s%d%s%d",s,&x,t,&y);
            if(t[0]=='p'&&!visL[x][y]) L[y]++,visL[x][y]=1;
            else if(!visR[x][y]) R[x]++,visR[x][y]=1;
        }
        ll cnt=0;
        for(int i=1;i<=M;i++){
            cnt+=L[i]*K+N*R[i]-L[i]*R[i];
        }
        cout<<N*M*K-cnt<<endl;
    }
    return 0;
}

A题：贪心。先比较两个的情况，推出贪心的依据，然后直接排序。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstdlib>

using namespace std;

const int maxn=1000100;

typedef long long ll;
int n;
const ll MOD=365*24*60*60;
struct NOde
{
    ll a,b;
    friend bool operator<(NOde A,NOde B)
    {
        return A.a*1.0*B.b<A.b*1.0*B.a;
    }
};
NOde Node[maxn];

int main()
{
   // freopen("in.txt","r",stdin);
    while(cin>>n&&n){
        for(int i=1;i<=n;i++){
            scanf("%lld%lld",&Node[i].a,&Node[i].b);
        }
        sort(Node+1,Node+n+1);
        ll res=0;
        for(int i=1;i<=n;i++){
            res=(res%MOD+((Node[i].a%MOD)+(Node[i].b%MOD)*(res%MOD))%MOD)%MOD;
        }
        cout<<res<<endl;
    }
    return 0;
}

D题：傻逼物理题。枚举角度，注意枚举的精度，少开变量，在纸上算出最终结果再写代码。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<math.h>

using namespace std;

const int maxn=1100;
const double Pi=acos(-1.0);
const double g=9.8;
const double EPS=0.0000000001;

typedef long long ll;

int N;
double H,L1,R1,L2,R2;
double v[maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>N&&N){
        cin>>H>>L1>>R1>>L2>>R2;
        for(int i=1;i<=N;i++) scanf("%lf",&v[i]);
        int ans=0;
        for(double a=0;a<=Pi/2;a+=0.0002){
            int cnt=0;
            for(int i=1;i<=N;i++){
                double x=v[i]*cos(a)*(v[i]*sin(a)+sqrt(v[i]*v[i]*sin(a)*sin(a)+2*g*H))/g;
                if(x>L2-EPS&&x<R2+EPS){
                    cnt=0;break;
                }
                if(x>L1-EPS&&x<R1+EPS) cnt++;
            }
            ans=max(cnt,ans);
        }
        for(double a=0;a<=Pi/2;a+=0.0002){
            int cnt=0;
            for(int i=1;i<=N;i++){
                double x=v[i]*cos(a)*(-v[i]*sin(a)+sqrt(v[i]*v[i]*sin(a)*sin(a)+2*g*H))/g;
                if(x>L2-EPS&&x<R2+EPS){
                    cnt=0;break;
                }
                if(x>L1-EPS&&x<R1+EPS) cnt++;
            }
            ans=max(cnt,ans);
        }
        cout<<ans<<endl;
    }
    return 0;
}

K题：傻逼模拟。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

const int maxn=1000100;

int N;
int x1,y1,x2,y2;
char d1,d2;
int s1,s2;
int t1,t2;
int K;

int goX(int x,char &dir,int s)
{
    if(dir=='S'){
        x=x+s;
        if(x>N){
            int c=x-N;
            int t=c/N;
            c%=N;
            if(t&1) x=c+1;
            else x=N-c,dir='N';
        }
        return x;
    }
    if(dir=='N'){
        x=x-s;
        if(x<1){
            int c=1-x;
            int t=c/N;
            c%=N;
            if(t&1) x=N-c;
            else x=c+1,dir='S';
        }
        return x;
    }
    return x;
}

int goY(int y,char &dir,int s)
{
    if(dir=='E'){
        y=y+s;
        if(y>N){
            int c=y-N;
            int t=c/N;
            c%=N;
            if(t&1) y=c+1;
            else y=N-c,dir='W';
        }
        return y;
    }
    if(dir=='W'){
        y=y-s;
        if(y<1){
            int c=1-y;
            int t=c/N;
            c%=N;
            if(t&1) y=N-c;
            else y=c+1,dir='E';
        }
        return y;
    }
    return y;
}

char Next(char dir)
{
    if(dir=='E') return 'N';
    if(dir=='N') return 'W';
    if(dir=='W') return 'S';
    if(dir=='S') return 'E';
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>N&&N){
        x1=y1=1;x2=y2=N;
        cin>>d1>>s1>>t1>>d2>>s2>>t2;
        cin>>K;
        int t=1;
        for(int i=1;i<=K;i++){
            //cout<<d1<<" "<<d2<<endl;
            x1=goX(x1,d1,s1);
            y1=goY(y1,d1,s1);
            x2=goX(x2,d2,s2);
            y2=goY(y2,d2,s2);
            //cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<" "<<d1<<" "<<d2<<endl;
            if(x1==x2&&y1==y2){
                swap(d1,d2);
                continue;
            }
            if(t%t1==0) d1=Next(d1);
            if(t%t2==0) d2=Next(d2);
            t++;
        }
        cout<<x1<<" "<<y1<<endl;
        cout<<x2<<" "<<y2<<endl;
    }
    return 0;
}