BUPT 2014新生暑假个人排位赛02

A. 丁神去谷歌

注意将除法改成交叉相乘

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath> using namespace std;  template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;} const int MAXN=10000000000;int n; int main(){    long long T,ma,mb,num=0;    read(T);    while(T--)    {        int a,b;        read(n);        read(ma),read(mb),num=1;        for(int i=1;i<n;i++)        {            read(a);read(b);            if(b*ma>a*mb)            {                ma=a;                mb=b;                num=i+1;            }            else if(b*ma==a*mb&&ma>a)            {                ma=a;                mb=b;                num=i+1;            }            else if(b*ma==a*mb&&ma==a&&num>i+1)            {                ma=a;                mb=b;                num=i+1;            }        }        printf("%lld\n",num);    }    return 0;}

B. 丁神又去谷歌

裸01背包

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath> using namespace std;  template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}//------------------------------------------------------------------- typedef long long ll; struct Node{    int ti,w;}a[110]; int n,t;ll dp[110][1010];  int main(){    int T;    read(T);    while(T--)    {        memset(dp,0,sizeof(dp));        read(t);read(n);        for(int i=1;i<=n;i++)        {            read(a[i].ti);read(a[i].w);        }        for(int i=1;i<=n;i++)        {            for(int j=0;j<=t;j++)            {                dp[i][j]=dp[i-1][j];                if(a[i].ti<=j)                    dp[i][j]=max(dp[i-1][j-a[i].ti]+a[i].w,dp[i][j]);            }        }         printf("%lld\n",dp[n][t]);    }    return 0;}

C. goblin 

数学题，排列组合。

首先，我们注意到这道题的某些公式，取模时候的除法，因为p不一定是质数，所以不一定能够用费马小定理求出逆元，因此一定要改变方法。

其次，我们可以发现这题是要求波浪数的个数，这种时候我比较推荐大家打表找规律，如形如/\/\/\/\/\（下称为A型）或者\/\/\/\/\/\/（下称为B型）的数字，当数字位数相同的时候，我们可以得知，A型数和B型数的个数是相同的，这一点我们可以通过逆序数证明，大家可以去百度一下证明方法。

那么，我们就可以将某些除以二的公式给略加改进

g[i]表示的是A型（也可以是B型，但只能是其中一个）的波浪数个数，用组合数学的插入法求出递推公式，注意，此时的g[i]只能当做一种类型来使用，因此每次记录都是j+=2的情况。

另外，如果数组都用long long 型，内存会爆，如果计算过程不用long long ，数据会爆。

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath> #define EPS 1e-10#define eps 1e-8 using namespace std;  template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}//------------------------------------------------------------------- const int MAXN=4300; int n,p;int g[MAXN];int c[MAXN][MAXN];  int main(){//    freopen("data.txt","r",stdin);    while(read(n)&&read(p))    {        memset(g,0,sizeof(g));        for(int i=1;i<=n;i++)            c[i][0]=1;        c[1][1]=1;        for(int i=2;i<=n;i++)        {            c[i][i]=1;            for(int j=1;j<i;j++)                c[i][j]=(c[i-1][j-1]%p+c[i-1][j]%p)%p;        }              g[0]=1%p;g[1]=1%p;g[2]=1%p;g[3]=2%p;        for(int i=4;i<=n;i++)            for(int j=0;j<=i-1;j+=2)            g[i]=(g[i]+((long long)c[i-1][j]%p*(long long)g[j]%p*(long long)g[i-1-j])%p)%p;        g[n]=(g[n]*2)%p;/*        int ans=0;        for(int i=0;i<n;i++)            a[i]=i+1;        do        {            int i;            for(i=1;i<n-1;i++)            {                if(a[i]>a[i-1]&&a[i]>a[i+1])                    continue;                else if(a[i]<a[i-1]&&a[i]<a[i+1])                    continue;                break;            }            if(i==n-1)                ans++;        }while(next_permutation(a,a+n));        cout<<n<<" : "<<ans<<endl;*/        cout<<g[n]<<endl;    }    return 0;}

D. 学姐逗学弟

裸的EVERY SG

自己去看看模板或者解法吧.......

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath> using namespace std;  template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}//------------------------------------------------------------------- const int MAXN=100010;int n,m;vector<int> a[MAXN];int sg[MAXN],po[MAXN];bool mex[MAXN];int step[MAXN]; void dfs(int x){    if(sg[x]!=-1)        return;    int sz=a[x].size(),maxx=-1*MAXN,minn=MAXN;    for(int i=0;i<sz;i++)    {        dfs(a[x][i]);        if(sg[a[x][i]]==0)            maxx=max(maxx,step[a[x][i]]);        minn=min(minn,step[a[x][i]]);    }    memset(mex,0,sizeof(bool)*(sz+10));    for(int i=0;i<sz;i++)        mex[sg[a[x][i]]]=true;    for(int i=0;i<sz+1;i++)        if(!mex[i])        {            sg[x]=i;            break;        }   if(sg[x]==0)        step[x]=minn+1;    else        step[x]=maxx+1;    if(sz==0)        step[x]=0;//    cout<<x<<" :"<<sg[x]<<endl;} void init(){    for(int i=0;i<MAXN;i++)        a[i].clear();    memset(sg,-1,sizeof(sg));    memset(po,0,sizeof(po));    memset(step,0,sizeof(step));} int main(){    int T;    read(T);    while(T--)    {        read(n);read(m);        init();        for(int i=2;i<=n;i++)        {            int temp;            read(temp);            a[temp].push_back(i);        }        for(int i=0;i<m;i++)            read(po[i]);        dfs(1);        int ans=-1;        for(int i=0;i<m;i++)        {            ans=max(step[po[i]],ans);//cout<<step[po[i]]<<" ";        }//    cout<<endl;        if(!(ans&1))            puts("So sad...");        else            puts("MengMengDa!");    }    return 0;}

E. 木头人足球赛

计算几何

大尧神代码

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>//#define LOCALusing namespace std;struct point{    double x,y,d,AngleUp,AngleDown;}Luke[20];struct PP{    double x,y;}UnCovered[20];const double eps=1e-7;const double pi=acos(-1.0);double xo,yo,AngleUp,AngleDown;int negativecnt;bool flag;bool GetLukeAngle(int x){    int i;    double d=sqrt((xo-Luke[x].x)*(xo-Luke[x].x)+(yo-Luke[x].y)*(yo-Luke[x].y));    //printf("d=%lf\n",d);    if(abs(Luke[x].d-d)<eps)return false;    double a=asin(Luke[x].d/d),b;    //printf("a=%lf\n",a);    double dy=Luke[x].y-yo,dx=Luke[x].x-xo;    //printf("dy=%lf dx=%lf\n",dy,dx);    b=atan2(dy,dx);    //printf("b=%lf\n",b);    if(b>eps)b=3*pi/2-b;    else if(b<-eps)b=-pi/2-b;    else b=pi*3/2;    Luke[x].AngleDown=b-a;    Luke[x].AngleUp=b+a;    //printf("Luke[%d].Angledown=%lf Luke[%d].AngleUp=%lf\n",x,Luke[x].AngleDown,x,Luke[x].AngleUp);    return true;}bool cmp(const point &a,const point &b){    if(abs(a.AngleDown-b.AngleDown)<eps)return a.AngleUp+eps<b.AngleUp;    return a.AngleDown+eps<b.AngleDown;}void Init(){    scanf("%lf%lf",&xo,&yo);//printf("xo=%lf yo=%lf\n",xo,yo);//cout<<xo<<' '<<yo<<endl;    AngleUp=atan2(38.000-yo,-xo);    if(AngleUp<-eps)AngleUp=-AngleUp-pi/2;    else if(AngleUp>eps) AngleUp=pi*3/2-AngleUp;    AngleDown=atan2(30.000-yo,-xo);    if(AngleDown<-eps)AngleDown=-AngleDown-pi/2;    else if(AngleDown>eps) AngleDown=pi*3/2-AngleDown;    //printf("Angledown=%lf AngleUp=%lf\n",AngleDown,AngleUp);    //cout<<AngleDown<<' '<<AngleUp<<endl;    flag=true;    for(int i=1;i<=11;i++)    {        scanf("%lf %lf %lf",&Luke[i].x,&Luke[i].y,&Luke[i].d);        if(flag==true)flag=GetLukeAngle(i);    }    //cout<<flag<<endl;}void Solve(){     sort(Luke+1,Luke+1+11,cmp);//for(int i=1;i<=11;i++)printf("Luke[%d].Angledown=%lf Luke[%d].AngleUp=%lf\n",i,Luke[i].AngleDown,i,Luke[i].AngleUp);    //for(int i=1;i<=11;i++)printf("Luke[%d].Angledown=%lf Luke[%d].AngleUp=%lf\n",i,Luke[i].AngleDown,i,Luke[i].AngleUp);    flag=false;    double start=AngleDown;    for(int i=1;i<=11;i++)    {            if(Luke[i].AngleUp+eps<AngleDown)continue;            if(Luke[i].AngleDown>AngleUp+eps)continue;            if(Luke[i].AngleDown>start+eps)            {                flag=true;break;            }            else if(Luke[i].AngleUp>start+eps)start=Luke[i].AngleUp;    }    if(flag==false&&start+eps<AngleUp)flag=true;}int main(){    #ifdef LOCAL    freopen("input.txt","r",stdin);    #endif // LOCAL    int T;    scanf("%d",&T);    //cout<<T<<endl;    while(T--)    {        Init();//if(T==6)printf("%d\n",flag);        if(flag==true)Solve();        if(flag)printf("Shoot!\n");        else printf("Poor Mays!\n");        //Solve();    }    return 0;}