Codeforces Beta Round #18 (Div. 2 Only)
http://codeforces.com/contest/18
A
暴力
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9 freopen("1.txt","r",stdin); 10 #endif */ 11 12 int a[6]; 13 14 void Check(char *s){ 15 int aa=sqr(a[0]-a[2])+sqr(a[1]-a[3]); 16 int bb=sqr(a[2]-a[4])+sqr(a[3]-a[5]); 17 int cc=sqr(a[4]-a[0])+sqr(a[5]-a[1]); 18 if((aa&&bb&&cc)==0) return; 19 // cout<<aa<<" "<<bb<<" "<<cc<<endl; 20 if(aa+bb==cc||aa+cc==bb||bb+cc==aa){ 21 cout<<s<<endl; 22 exit(0); 23 } 24 } 25 26 int main(){ 27 #ifndef ONLINE_JUDGE 28 // freopen("1.txt","r",stdin); 29 #endif 30 // std::ios::sync_with_stdio(false); 31 for(int i=0;i<6;i++) cin>>a[i]; 32 Check("RIGHT"); 33 for(int i=0;i<6;i++){ 34 // cout<<a[i]<<"g"<<endl; 35 a[i]--; 36 Check("ALMOST"); 37 a[i]+=2; 38 Check("ALMOST"); 39 a[i]--; 40 // cout<<a[i]<<"h"<<endl; 41 42 } 43 cout<<"NEITHER"<<endl; 44 45 }
B
水题
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9 freopen("1.txt","r",stdin); 10 #endif */ 11 12 13 int main(){ 14 #ifndef ONLINE_JUDGE 15 // freopen("1.txt","r",stdin); 16 #endif 17 std::ios::sync_with_stdio(false); 18 int n,m,l,d; 19 cin>>n>>d>>m>>l; 20 ll x=0,L=-m,R=-m+l; 21 for(int i=0;i<n;i++){ 22 L+=m; 23 R+=m; 24 if(x+d<L) break; 25 ll o=(R-x)/d; 26 x+=d*o; 27 } 28 cout<<x+d<<endl; 29 }
C
水题
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #include<iostream> 5 using namespace std; 6 typedef long long ll; 7 8 int n; 9 ll a[100005]; 10 11 int main(){ 12 cin>>n; 13 ll ans=0; 14 int b; 15 for(int i=1;i<=n;i++){ 16 cin>>b; 17 a[i]=a[i-1]+b; 18 } 19 for(int i=1;i<=n-1;i++){ 20 if(a[n]-a[i]==a[i]) ans++; 21 } 22 cout<<ans<<endl; 23 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9 freopen("1.txt","r",stdin); 10 #endif */ 11 12 #define MAXN 9999 13 #define MAXSIZE 5007//位数 14 #define DLEN 4 15 16 class BigNum 17 { 18 private: 19 int a[5007]; //可以控制大数的位数 20 int len; //大数长度 21 public: 22 BigNum() 23 { 24 len = 1; //构造函数 25 memset(a,0,sizeof(a)); 26 } 27 BigNum(const int); //将一个int类型的变量转化为大数 28 BigNum(const char*); //将一个字符串类型的变量转化为大数 29 BigNum(const BigNum &); //拷贝构造函数 30 BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算 31 32 friend istream& operator>>(istream&, BigNum&); //重载输入运算符 33 friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符 34 35 BigNum operator+(const BigNum &) const; //重载加法运算符，两个大数之间的相加运算 36 BigNum operator-(const BigNum &) const; //重载减法运算符，两个大数之间的相减运算 37 BigNum operator*(const BigNum &) const; //重载乘法运算符，两个大数之间的相乘运算 38 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 39 40 BigNum operator^(const int &) const; //大数的n次方运算 41 int operator%(const int &) const; //大数对一个int类型的变量进行取模运算 42 bool operator>(const BigNum & T)const; //大数和另一个大数的大小比较 43 bool operator>(const int & t)const; //大数和一个int类型的变量的大小比较 44 45 void print(); //输出大数 46 }; 47 BigNum::BigNum(const int b) //将一个int类型的变量转化为大数 48 { 49 int c,d = b; 50 len = 0; 51 memset(a,0,sizeof(a)); 52 while(d > MAXN) 53 { 54 c = d - (d / (MAXN + 1)) * (MAXN + 1); 55 d = d / (MAXN + 1); 56 a[len++] = c; 57 } 58 a[len++] = d; 59 } 60 BigNum::BigNum(const char*s) //将一个字符串类型的变量转化为大数 61 { 62 int t,k,index,l,i; 63 memset(a,0,sizeof(a)); 64 l=strlen(s); 65 len=l/DLEN; 66 if(l%DLEN) 67 len++; 68 index=0; 69 for(i=l-1; i>=0; i-=DLEN) 70 { 71 t=0; 72 k=i-DLEN+1; 73 if(k<0) 74 k=0; 75 for(int j=k; j<=i; j++) 76 t=t*10+s[j]-'0'; 77 a[index++]=t; 78 } 79 } 80 BigNum::BigNum(const BigNum & T) : len(T.len) //拷贝构造函数 81 { 82 int i; 83 memset(a,0,sizeof(a)); 84 for(i = 0 ; i < len ; i++) 85 a[i] = T.a[i]; 86 } 87 BigNum & BigNum::operator=(const BigNum & n) //重载赋值运算符，大数之间进行赋值运算 88 { 89 int i; 90 len = n.len; 91 memset(a,0,sizeof(a)); 92 for(i = 0 ; i < len ; i++) 93 a[i] = n.a[i]; 94 return *this; 95 } 96 istream& operator>>(istream & in, BigNum & b) //重载输入运算符 97 { 98 char ch[MAXSIZE*4]; 99 int i = -1; 100 in>>ch; 101 int l=strlen(ch); 102 int count=0,sum=0; 103 for(i=l-1; i>=0;) 104 { 105 sum = 0; 106 int t=1; 107 for(int j=0; j<4&&i>=0; j++,i--,t*=10) 108 { 109 sum+=(ch[i]-'0')*t; 110 } 111 b.a[count]=sum; 112 count++; 113 } 114 b.len =count++; 115 return in; 116 117 } 118 ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符 119 { 120 int i; 121 cout << b.a[b.len - 1]; 122 for(i = b.len - 2 ; i >= 0 ; i--) 123 { 124 cout.width(DLEN); 125 cout.fill('0'); 126 cout << b.a[i]; 127 } 128 return out; 129 } 130 131 BigNum BigNum::operator+(const BigNum & T) const //两个大数之间的相加运算 132 { 133 BigNum t(*this); 134 int i,big; //位数 135 big = T.len > len ? T.len : len; 136 for(i = 0 ; i < big ; i++) 137 { 138 t.a[i] +=T.a[i]; 139 if(t.a[i] > MAXN) 140 { 141 t.a[i + 1]++; 142 t.a[i] -=MAXN+1; 143 } 144 } 145 if(t.a[big] != 0) 146 t.len = big + 1; 147 else 148 t.len = big; 149 return t; 150 } 151 BigNum BigNum::operator-(const BigNum & T) const //两个大数之间的相减运算 152 { 153 int i,j,big; 154 bool flag; 155 BigNum t1,t2; 156 if(*this>T) 157 { 158 t1=*this; 159 t2=T; 160 flag=0; 161 } 162 else 163 { 164 t1=T; 165 t2=*this; 166 flag=1; 167 } 168 big=t1.len; 169 for(i = 0 ; i < big ; i++) 170 { 171 if(t1.a[i] < t2.a[i]) 172 { 173 j = i + 1; 174 while(t1.a[j] == 0) 175 j++; 176 t1.a[j--]--; 177 while(j > i) 178 t1.a[j--] += MAXN; 179 t1.a[i] += MAXN + 1 - t2.a[i]; 180 } 181 else 182 t1.a[i] -= t2.a[i]; 183 } 184 t1.len = big; 185 while(t1.a[t1.len - 1] == 0 && t1.len > 1) 186 { 187 t1.len--; 188 big--; 189 } 190 if(flag) 191 t1.a[big-1]=0-t1.a[big-1]; 192 return t1; 193 } 194 195 BigNum BigNum::operator*(const BigNum & T) const //两个大数之间的相乘运算 196 { 197 BigNum ret; 198 int i,j,up; 199 int temp,temp1; 200 for(i = 0 ; i < len ; i++) 201 { 202 up = 0; 203 for(j = 0 ; j < T.len ; j++) 204 { 205 temp = a[i] * T.a[j] + ret.a[i + j] + up; 206 if(temp > MAXN) 207 { 208 temp1 = temp - temp / (MAXN + 1) * (MAXN + 1); 209 up = temp / (MAXN + 1); 210 ret.a[i + j] = temp1; 211 } 212 else 213 { 214 up = 0; 215 ret.a[i + j] = temp; 216 } 217 } 218 if(up != 0) 219 ret.a[i + j] = up; 220 } 221 ret.len = i + j; 222 while(ret.a[ret.len - 1] == 0 && ret.len > 1) 223 ret.len--; 224 return ret; 225 } 226 BigNum BigNum::operator/(const int & b) const //大数对一个整数进行相除运算 227 { 228 BigNum ret; 229 int i,down = 0; 230 for(i = len - 1 ; i >= 0 ; i--) 231 { 232 ret.a[i] = (a[i] + down * (MAXN + 1)) / b; 233 down = a[i] + down * (MAXN + 1) - ret.a[i] * b; 234 } 235 ret.len = len; 236 while(ret.a[ret.len - 1] == 0 && ret.len > 1) 237 ret.len--; 238 return ret; 239 } 240 int BigNum::operator %(const int & b) const //大数对一个int类型的变量进行取模运算 241 { 242 int i,d=0; 243 for (i = len-1; i>=0; i--) 244 { 245 d = ((d * (MAXN+1))% b + a[i])% b; 246 } 247 return d; 248 } 249 BigNum BigNum::operator^(const int & n) const //大数的n次方运算 250 { 251 BigNum t,ret(1); 252 int i; 253 if(n<0) 254 exit(-1); 255 if(n==0) 256 return 1; 257 if(n==1) 258 return *this; 259 int m=n; 260 while(m>1) 261 { 262 t=*this; 263 for( i=1; i<<1<=m; i<<=1) 264 { 265 t=t*t; 266 } 267 m-=i; 268 ret=ret*t; 269 if(m==1) 270 ret=ret*(*this); 271 } 272 return ret; 273 } 274 bool BigNum::operator>(const BigNum & T) const //大数和另一个大数的大小比较 275 { 276 int ln; 277 if(len > T.len) 278 return true; 279 else if(len == T.len) 280 { 281 ln = len - 1; 282 while(a[ln] == T.a[ln] && ln >= 0) 283 ln--; 284 if(ln >= 0 && a[ln] > T.a[ln]) 285 return true; 286 else 287 return false; 288 } 289 else 290 return false; 291 } 292 bool BigNum::operator >(const int & t) const //大数和一个int类型的变量的大小比较 293 { 294 BigNum b(t); 295 return *this>b; 296 } 297 298 void BigNum::print() //输出大数 299 { 300 int i; 301 cout << a[len - 1]; 302 for(i = len - 2 ; i >= 0 ; i--) 303 { 304 cout.width(DLEN); 305 cout.fill('0'); 306 cout << a[i]; 307 } 308 cout << endl; 309 } 310 BigNum num1[5007]; 311 char str[5007][10]; 312 int num2[5007],num3[5007]; 313 BigNum Jud(int n,int m) 314 { 315 BigNum sum,mul; 316 sum=BigNum(1); 317 mul=BigNum(2); 318 while(m) 319 { 320 if(m&1) 321 sum=sum*mul; 322 mul=mul*mul; 323 m>>=1; 324 } 325 return sum; 326 } 327 328 int n; 329 struct sair{ 330 string str; 331 int num; 332 int pos; 333 }a[5005]; 334 335 int book[5005]; 336 337 bool cmp(sair a,sair b){ 338 return a.num==b.num?a.pos>b.pos:a.num>b.num; 339 } 340 341 bool Check(int x,int y){ 342 for(int i=x;i<=y;i++){ 343 if(book[i]) return false; 344 } 345 for(int i=x;i<=y;i++){ 346 book[i]=1; 347 } 348 return true; 349 } 350 351 int main(){ 352 #ifndef ONLINE_JUDGE 353 freopen("1.txt","r",stdin); 354 #endif 355 //std::ios::sync_with_stdio(false); 356 cin>>n; 357 for(int i=1;i<=n;i++){ 358 cin>>a[i].str>>a[i].num; 359 a[i].pos=i; 360 } 361 sort(a+1,a+n+1,cmp); 362 vector<int>ve; 363 for(int i=1;i<=n;i++){ 364 if(a[i].str=="sell"){ 365 for(int j=i+1;j<=n;j++){ 366 if(a[j].str=="win"&&a[j].num==a[i].num){ 367 if(Check(a[j].pos,a[i].pos)){ 368 ve.push_back(a[i].num); 369 } 370 } 371 } 372 } 373 } 374 BigNum ans=0; 375 for(int i=0;i<ve.size();i++){ 376 ans=ans+Jud(2,ve[i]); 377 } 378 ans.print(); 379 380 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 505 7 #define maxc 35 8 typedef long long ll; 9 /*#ifndef ONLINE_JUDGE 10 freopen("1.txt","r",stdin); 11 #endif */ 12 13 int n,m; 14 int dp[maxn][maxc][maxc]; 15 char color[maxn][maxn]; 16 int cost[maxn][maxc][maxc]; 17 pair<int,int>pre[maxn][maxc][maxc]; 18 19 void print(int floor,int ii,int jj){ 20 if(floor>1){ 21 print(floor-1,pre[floor][ii][jj].first,pre[floor][ii][jj].second); 22 } 23 for(int i=1;i<=m;i++){ 24 if(i&1) printf("%c",'a'-1+ii); 25 else printf("%c",'a'-1+jj); 26 } 27 printf("\n"); 28 } 29 30 int main(){ 31 #ifndef ONLINE_JUDGE 32 freopen("1.txt","r",stdin); 33 #endif 34 //std::ios::sync_with_stdio(false); 35 scanf("%d %d",&n,&m); 36 for(int i=1;i<=n;i++){ 37 scanf("%s",color[i]+1); 38 } 39 memset(dp,0x3f,sizeof(dp)); 40 memset(cost,0x3f,sizeof(dp)); 41 for(int i=1;i<=n;i++){ 42 for(int j=1;j<=26;j++){ 43 for(int k=1;k<=26;k++){ 44 if(j!=k){ 45 cost[i][j][k]=0; 46 for(int l=1;l<=m;l+=2){ 47 cost[i][j][k]+=(color[i][l]!=('a'-1+j)); 48 } 49 for(int l=2;l<=m;l+=2){ 50 cost[i][j][k]+=(color[i][l]!=('a'-1+k)); 51 } 52 } 53 } 54 } 55 } 56 for(int i=1;i<=26;i++){ 57 for(int j=1;j<=26;j++){ 58 dp[1][i][j]=cost[1][i][j]; 59 } 60 } 61 for(int i=1;i<=n;i++){ 62 for(int j=1;j<=26;j++){ 63 for(int k=1;k<=26;k++){ 64 for(int p=1;p<=26;p++){ 65 for(int q=1;q<=26;q++){ 66 if(p!=j&&q!=k){ 67 if(dp[i][j][k]>dp[i-1][p][q]+cost[i][j][k]){ 68 pre[i][j][k]=make_pair(p,q); 69 dp[i][j][k]=dp[i-1][p][q]+cost[i][j][k]; 70 } 71 } 72 } 73 } 74 } 75 } 76 } 77 int ans=0x3f3f3f3f; 78 int ansi,ansj; 79 for(int i=1;i<=26;i++){ 80 for(int j=1;j<=26;j++){ 81 if(ans>dp[n][i][j]){ 82 ans=dp[n][ansi=i][ansj=j]; 83 } 84 } 85 } 86 printf("%d\n",ans); 87 print(n,ansi,ansj); 88 }