Codeforces Beta Round #18 (Div. 2 Only)

Codeforces Beta Round #18 (Div. 2 Only)

http://codeforces.com/contest/18

A

暴力

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define maxn 500005
typedef long long ll;
/*#ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
#endif */

int a[6];

void Check(char *s){
    int aa=sqr(a[0]-a[2])+sqr(a[1]-a[3]);
    int bb=sqr(a[2]-a[4])+sqr(a[3]-a[5]);
    int cc=sqr(a[4]-a[0])+sqr(a[5]-a[1]);
    if((aa&&bb&&cc)==0) return;
  //  cout<<aa<<" "<<bb<<" "<<cc<<endl;
    if(aa+bb==cc||aa+cc==bb||bb+cc==aa){
        cout<<s<<endl;
        exit(0);
    }
}

int main(){
    #ifndef ONLINE_JUDGE
       // freopen("1.txt","r",stdin);
    #endif
   // std::ios::sync_with_stdio(false);
    for(int i=0;i<6;i++) cin>>a[i];
    Check("RIGHT");
    for(int i=0;i<6;i++){
      //  cout<<a[i]<<"g"<<endl;
        a[i]--;
        Check("ALMOST");
        a[i]+=2;
        Check("ALMOST");
        a[i]--;
     //   cout<<a[i]<<"h"<<endl;

    }
    cout<<"NEITHER"<<endl;

}

B

水题

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define maxn 500005
typedef long long ll;
/*#ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
#endif */


int main(){
    #ifndef ONLINE_JUDGE
  //      freopen("1.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    int n,m,l,d;
    cin>>n>>d>>m>>l;
    ll x=0,L=-m,R=-m+l;
    for(int i=0;i<n;i++){
        L+=m;
        R+=m;
        if(x+d<L) break;
        ll o=(R-x)/d;
        x+=d*o;
    }
    cout<<x+d<<endl;
}

C

水题

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;

int n;
ll a[100005]; 
 
int main(){
    cin>>n;
    ll ans=0;
    int b;
    for(int i=1;i<=n;i++){
        cin>>b;
        a[i]=a[i-1]+b;
    }
    for(int i=1;i<=n-1;i++){
        if(a[n]-a[i]==a[i]) ans++;
    }
    cout<<ans<<endl;
}

D

贪心，因为有2^n>2^(n-1)+2^(n-2)+...+1的规律，所以可以优先挑选大的。挑选的时候是从后往前选比较好写

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define maxn 500005
typedef long long ll;
/*#ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
#endif */

#define MAXN 9999
#define MAXSIZE 5007//位数
#define DLEN 4

class BigNum
{
private:
    int a[5007];    //可以控制大数的位数
    int len;       //大数长度
public:
    BigNum()
    {
        len = 1;    //构造函数
        memset(a,0,sizeof(a));
    }
    BigNum(const int);       //将一个int类型的变量转化为大数
    BigNum(const char*);     //将一个字符串类型的变量转化为大数
    BigNum(const BigNum &);  //拷贝构造函数
    BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算

    friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符
    friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符

    BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算
    BigNum operator-(const BigNum &) const;   //重载减法运算符，两个大数之间的相减运算
    BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算
    BigNum operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算

    BigNum operator^(const int  &) const;    //大数的n次方运算
    int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算
    bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较
    bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较

    void print();       //输出大数
};
BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数
{
    int c,d = b;
    len = 0;
    memset(a,0,sizeof(a));
    while(d > MAXN)
    {
        c = d - (d / (MAXN + 1)) * (MAXN + 1);
        d = d / (MAXN + 1);
        a[len++] = c;
    }
    a[len++] = d;
}
BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数
{
    int t,k,index,l,i;
    memset(a,0,sizeof(a));
    l=strlen(s);
    len=l/DLEN;
    if(l%DLEN)
        len++;
    index=0;
    for(i=l-1; i>=0; i-=DLEN)
    {
        t=0;
        k=i-DLEN+1;
        if(k<0)
            k=0;
        for(int j=k; j<=i; j++)
            t=t*10+s[j]-'0';
        a[index++]=t;
    }
}
BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
{
    int i;
    memset(a,0,sizeof(a));
    for(i = 0 ; i < len ; i++)
        a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算
{
    int i;
    len = n.len;
    memset(a,0,sizeof(a));
    for(i = 0 ; i < len ; i++)
        a[i] = n.a[i];
    return *this;
}
istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符
{
    char ch[MAXSIZE*4];
    int i = -1;
    in>>ch;
    int l=strlen(ch);
    int count=0,sum=0;
    for(i=l-1; i>=0;)
    {
        sum = 0;
        int t=1;
        for(int j=0; j<4&&i>=0; j++,i--,t*=10)
        {
            sum+=(ch[i]-'0')*t;
        }
        b.a[count]=sum;
        count++;
    }
    b.len =count++;
    return in;

}
ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符
{
    int i;
    cout << b.a[b.len - 1];
    for(i = b.len - 2 ; i >= 0 ; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << b.a[i];
    }
    return out;
}

BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算
{
    BigNum t(*this);
    int i,big;      //位数
    big = T.len > len ? T.len : len;
    for(i = 0 ; i < big ; i++)
    {
        t.a[i] +=T.a[i];
        if(t.a[i] > MAXN)
        {
            t.a[i + 1]++;
            t.a[i] -=MAXN+1;
        }
    }
    if(t.a[big] != 0)
        t.len = big + 1;
    else
        t.len = big;
    return t;
}
BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算
{
    int i,j,big;
    bool flag;
    BigNum t1,t2;
    if(*this>T)
    {
        t1=*this;
        t2=T;
        flag=0;
    }
    else
    {
        t1=T;
        t2=*this;
        flag=1;
    }
    big=t1.len;
    for(i = 0 ; i < big ; i++)
    {
        if(t1.a[i] < t2.a[i])
        {
            j = i + 1;
            while(t1.a[j] == 0)
                j++;
            t1.a[j--]--;
            while(j > i)
                t1.a[j--] += MAXN;
            t1.a[i] += MAXN + 1 - t2.a[i];
        }
        else
            t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while(t1.a[t1.len - 1] == 0 && t1.len > 1)
    {
        t1.len--;
        big--;
    }
    if(flag)
        t1.a[big-1]=0-t1.a[big-1];
    return t1;
}

BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算
{
    BigNum ret;
    int i,j,up;
    int temp,temp1;
    for(i = 0 ; i < len ; i++)
    {
        up = 0;
        for(j = 0 ; j < T.len ; j++)
        {
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if(temp > MAXN)
            {
                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                up = temp / (MAXN + 1);
                ret.a[i + j] = temp1;
            }
            else
            {
                up = 0;
                ret.a[i + j] = temp;
            }
        }
        if(up != 0)
            ret.a[i + j] = up;
    }
    ret.len = i + j;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{
    BigNum ret;
    int i,down = 0;
    for(i = len - 1 ; i >= 0 ; i--)
    {
        ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
    }
    ret.len = len;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算
{
    int i,d=0;
    for (i = len-1; i>=0; i--)
    {
        d = ((d * (MAXN+1))% b + a[i])% b;
    }
    return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{
    BigNum t,ret(1);
    int i;
    if(n<0)
        exit(-1);
    if(n==0)
        return 1;
    if(n==1)
        return *this;
    int m=n;
    while(m>1)
    {
        t=*this;
        for( i=1; i<<1<=m; i<<=1)
        {
            t=t*t;
        }
        m-=i;
        ret=ret*t;
        if(m==1)
            ret=ret*(*this);
    }
    return ret;
}
bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较
{
    int ln;
    if(len > T.len)
        return true;
    else if(len == T.len)
    {
        ln = len - 1;
        while(a[ln] == T.a[ln] && ln >= 0)
            ln--;
        if(ln >= 0 && a[ln] > T.a[ln])
            return true;
        else
            return false;
    }
    else
        return false;
}
bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较
{
    BigNum b(t);
    return *this>b;
}

void BigNum::print()    //输出大数
{
    int i;
    cout << a[len - 1];
    for(i = len - 2 ; i >= 0 ; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << a[i];
    }
    cout << endl;
}
BigNum num1[5007];
char str[5007][10];
int num2[5007],num3[5007];
BigNum Jud(int n,int m)
{
    BigNum sum,mul;
    sum=BigNum(1);
    mul=BigNum(2);
    while(m)
    {
        if(m&1)
         sum=sum*mul;
         mul=mul*mul;
         m>>=1;
    }
    return sum;
}

int n;
struct sair{
    string str;
    int num;
    int pos;
}a[5005];

int book[5005];

bool cmp(sair a,sair b){
    return a.num==b.num?a.pos>b.pos:a.num>b.num;
}

bool Check(int x,int y){
    for(int i=x;i<=y;i++){
        if(book[i]) return false;
    }
    for(int i=x;i<=y;i++){
        book[i]=1;
    }
    return true;
}

int main(){
    #ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
    #endif
    //std::ios::sync_with_stdio(false);
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i].str>>a[i].num;
        a[i].pos=i;
    }
    sort(a+1,a+n+1,cmp);
    vector<int>ve;
    for(int i=1;i<=n;i++){
        if(a[i].str=="sell"){
            for(int j=i+1;j<=n;j++){
                if(a[j].str=="win"&&a[j].num==a[i].num){
                    if(Check(a[j].pos,a[i].pos)){
                        ve.push_back(a[i].num);
                    }
                }
            }
        }
    }
    BigNum ans=0;
    for(int i=0;i<ve.size();i++){
        ans=ans+Jud(2,ve[i]);
    }
    ans.print();

}

E

dp

先预处理出每个格子各个颜色的花费，然后开始DP

参考博客：https://blog.csdn.net/c20190102/article/details/81775186

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define maxn 505
#define maxc 35
typedef long long ll;
/*#ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
#endif */

int n,m;
int dp[maxn][maxc][maxc];
char color[maxn][maxn];
int cost[maxn][maxc][maxc];
pair<int,int>pre[maxn][maxc][maxc];

void print(int floor,int ii,int jj){
    if(floor>1){
        print(floor-1,pre[floor][ii][jj].first,pre[floor][ii][jj].second);
    }
    for(int i=1;i<=m;i++){
        if(i&1) printf("%c",'a'-1+ii);
        else printf("%c",'a'-1+jj);
    }
    printf("
");
}

int main(){
    #ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
    #endif
    //std::ios::sync_with_stdio(false);
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%s",color[i]+1);
    }
    memset(dp,0x3f,sizeof(dp));
    memset(cost,0x3f,sizeof(dp));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=26;j++){
            for(int k=1;k<=26;k++){
                if(j!=k){
                    cost[i][j][k]=0;
                    for(int l=1;l<=m;l+=2){
                        cost[i][j][k]+=(color[i][l]!=('a'-1+j));
                    }
                    for(int l=2;l<=m;l+=2){
                        cost[i][j][k]+=(color[i][l]!=('a'-1+k));
                    }
                }
            }
        }
    }
    for(int i=1;i<=26;i++){
        for(int j=1;j<=26;j++){
            dp[1][i][j]=cost[1][i][j];
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=26;j++){
            for(int k=1;k<=26;k++){
                for(int p=1;p<=26;p++){
                    for(int q=1;q<=26;q++){
                        if(p!=j&&q!=k){
                            if(dp[i][j][k]>dp[i-1][p][q]+cost[i][j][k]){
                                pre[i][j][k]=make_pair(p,q);
                                dp[i][j][k]=dp[i-1][p][q]+cost[i][j][k];
                            }
                        }
                    }
                }
            }
        }
    }
    int ans=0x3f3f3f3f;
    int ansi,ansj;
    for(int i=1;i<=26;i++){
        for(int j=1;j<=26;j++){
            if(ans>dp[n][i][j]){
                ans=dp[n][ansi=i][ansj=j];
            }
        }
    }
    printf("%d
",ans);
    print(n,ansi,ansj);
}