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[];

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

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

 int main(){
     cin>>n;
     ll ans=;
     int b;
     for(int i=;i<=n;i++){
         cin>>b;
         a[i]=a[i-]+b;
     }
     for(int i=;i<=n-;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[];    //可以控制大数的位数
     int len;       //大数长度
 public:
     BigNum()
     {
         len = ;    //构造函数
         memset(a,,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 = ;
     memset(a,,sizeof(a));
     while(d > MAXN)
     {
         c = d - (d / (MAXN + )) * (MAXN + );
         d = d / (MAXN + );
         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,,sizeof(a));
     l=strlen(s);
     len=l/DLEN;
     if(l%DLEN)
         len++;
     index=;
     for(i=l-; i>=; i-=DLEN)
     {
         t=;
         k=i-DLEN+;
         if(k<)
             k=;
         for(int j=k; j<=i; j++)
             t=t*+s[j]-'';
         a[index++]=t;
     }
 }
 BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
 {
     int i;
     memset(a,,sizeof(a));
     for(i =  ; 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,,sizeof(a));
     for(i =  ; 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*];
     int i = -;
     in>>ch;
     int l=strlen(ch);
     int count=,sum=;
     for(i=l-; i>=;)
     {
         sum = ;
         int t=;
         for(int j=; j<&&i>=; j++,i--,t*=)
         {
             sum+=(ch[i]-'')*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 - ];
     for(i = b.len -  ; i >=  ; i--)
     {
         cout.width(DLEN);
         cout.fill('');
         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 =  ; i < big ; i++)
     {
         t.a[i] +=T.a[i];
         if(t.a[i] > MAXN)
         {
             t.a[i + ]++;
             t.a[i] -=MAXN+;
         }
     }
     if(t.a[big] != )
         t.len = big + ;
     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=;
     }
     else
     {
         t1=T;
         t2=*this;
         flag=;
     }
     big=t1.len;
     for(i =  ; i < big ; i++)
     {
         if(t1.a[i] < t2.a[i])
         {
             j = i + ;
             while(t1.a[j] == )
                 j++;
             t1.a[j--]--;
             while(j > i)
                 t1.a[j--] += MAXN;
             t1.a[i] += MAXN +  - t2.a[i];
         }
         else
             t1.a[i] -= t2.a[i];
     }
     t1.len = big;
     while(t1.a[t1.len - ] ==  && t1.len > )
     {
         t1.len--;
         big--;
     }
     if(flag)
         t1.a[big-]=-t1.a[big-];
     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 =  ; i < len ; i++)
     {
         up = ;
         for(j =  ; j < T.len ; j++)
         {
             temp = a[i] * T.a[j] + ret.a[i + j] + up;
             if(temp > MAXN)
             {
                 temp1 = temp - temp / (MAXN + ) * (MAXN + );
                 up = temp / (MAXN + );
                 ret.a[i + j] = temp1;
             }
             else
             {
                 up = ;
                 ret.a[i + j] = temp;
             }
         }
         if(up != )
             ret.a[i + j] = up;
     }
     ret.len = i + j;
     while(ret.a[ret.len - ] ==  && ret.len > )
         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 = ;
     for(i = len -  ; i >=  ; i--)
     {
         ret.a[i] = (a[i] + down * (MAXN + )) / b;
         down = a[i] + down * (MAXN + ) - ret.a[i] * b;
     }
     ret.len = len;
     while(ret.a[ret.len - ] ==  && ret.len > )
         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=;
     for (i = len-; i>=; i--)
     {
         d = ((d * (MAXN+))% b + a[i])% b;
     }
     return d;
 }
 BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
 {
     BigNum t,ret();
     int i;
     if(n<)
         exit(-);
     if(n==)
         return ;
     if(n==)
         return *this;
     int m=n;
     while(m>)
     {
         t=*this;
         for( i=; i<<<=m; i<<=)
         {
             t=t*t;
         }
         m-=i;
         ret=ret*t;
         if(m==)
             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 - ;
         while(a[ln] == T.a[ln] && ln >= )
             ln--;
         if(ln >=  && 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 - ];
     for(i = len -  ; i >=  ; i--)
     {
         cout.width(DLEN);
         cout.fill('');
         cout << a[i];
     }
     cout << endl;
 }
 BigNum num1[];
 char str[][];
 int num2[],num3[];
 BigNum Jud(int n,int m)
 {
     BigNum sum,mul;
     sum=BigNum();
     mul=BigNum();
     while(m)
     {
         if(m&)
          sum=sum*mul;
          mul=mul*mul;
          m>>=;
     }
     return sum;
 }

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

 int book[];

 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]=;
     }
     return true;
 }

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
     #endif
     //std::ios::sync_with_stdio(false);
     cin>>n;
     for(int i=;i<=n;i++){
         cin>>a[i].str>>a[i].num;
         a[i].pos=i;
     }
     sort(a+,a+n+,cmp);
     vector<int>ve;
     for(int i=;i<=n;i++){
         if(a[i].str=="sell"){
             for(int j=i+;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=;
     for(int i=;i<ve.size();i++){
         ans=ans+Jud(,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>){
         print(floor-,pre[floor][ii][jj].first,pre[floor][ii][jj].second);
     }
     for(int i=;i<=m;i++){
         if(i&) printf("%c",'a'-+ii);
         else printf("%c",'a'-+jj);
     }
     printf("\n");
 }

 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=;i<=n;i++){
         scanf("%s",color[i]+);
     }
     memset(dp,0x3f,sizeof(dp));
     memset(cost,0x3f,sizeof(dp));
     for(int i=;i<=n;i++){
         for(int j=;j<=;j++){
             for(int k=;k<=;k++){
                 if(j!=k){
                     cost[i][j][k]=;
                     for(int l=;l<=m;l+=){
                         cost[i][j][k]+=(color[i][l]!=('a'-+j));
                     }
                     for(int l=;l<=m;l+=){
                         cost[i][j][k]+=(color[i][l]!=('a'-+k));
                     }
                 }
             }
         }
     }
     for(int i=;i<=;i++){
         for(int j=;j<=;j++){
             dp[][i][j]=cost[][i][j];
         }
     }
     for(int i=;i<=n;i++){
         for(int j=;j<=;j++){
             for(int k=;k<=;k++){
                 for(int p=;p<=;p++){
                     for(int q=;q<=;q++){
                         if(p!=j&&q!=k){
                             if(dp[i][j][k]>dp[i-][p][q]+cost[i][j][k]){
                                 pre[i][j][k]=make_pair(p,q);
                                 dp[i][j][k]=dp[i-][p][q]+cost[i][j][k];
                             }
                         }
                     }
                 }
             }
         }
     }
     int ans=0x3f3f3f3f;
     int ansi,ansj;
     for(int i=;i<=;i++){
         for(int j=;j<=;j++){
             if(ans>dp[n][i][j]){
                 ans=dp[n][ansi=i][ansj=j];
             }
         }
     }
     printf("%d\n",ans);
     print(n,ansi,ansj);
 }