[Codeforces Gym 101243][2016acm/icpc Quarterfinal Central region of Russia]

来源：互联网发布：杨颖开的淘宝店铺名字编辑：程序博客网时间：2024/05/29 11:22

传送门
诶比赛打得贼爽，遇到了travel poorly
最后以一吨罚时，10/11题提前两小时结束比赛滚回宿舍了
随便整理一下题目吧
A
题意：需要烤n(n<=500)条鱼，每次能烤k条鱼的一面，询问至少需要烤多少次才能烤完所有的鱼。（每条鱼的两面都需要被烤）
解答：直接输出2nk，和2取最大值，向上取整。

#include <bits/stdc++.h>using namespace std;int main() {    freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    int n,k; cin >> n >> k;    cout << max( 2, (int)ceil( (double)2*(double)n/(double)k ) );    return 0;}

B
题意：经典汉诺塔操作，询问在进行到多少步的时候第一次出现三个柱子上的圆盘数量相等，圆盘数n<=300
解答：打表找规律
奇数项推偶数项： f(x)=4f(x−1)+4f(x−2)−16f(x−3)−21
偶数项推奇数项： f(x)=4f(x−1)+2
诶没想到这么奇葩的数列规律也能被我们找到
剩下的就是高精度了

#include <iostream>  #include <cstdio>  #include <cstdlib>  #include <cstring>  #include <string>  #include <algorithm>  using namespace std;  const int MAXN = 410;  struct bign  {      int len, s[MAXN];      bign ()      {          memset(s, 0, sizeof(s));          len = 1;      }      bign (int num) { *this = num; }      bign (const char *num) { *this = num; }      bign operator = (const int num)      {          char s[MAXN];          sprintf(s, "%d", num);          *this = s;          return *this;      }      bign operator = (const char *num)      {          for(int i = 0; num[i] == '0'; num++) ;  //È¥Ç°µ¼0          len = strlen(num);          for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';          return *this;      }      bign operator + (const bign &b) const //+      {          bign c;          c.len = 0;          for(int i = 0, g = 0; g || i < max(len, b.len); i++)          {              int x = g;              if(i < len) x += s[i];              if(i < b.len) x += b.s[i];              c.s[c.len++] = x % 10;              g = x / 10;          }          return c;      }      bign operator += (const bign &b)      {          *this = *this + b;          return *this;      }      void clean()      {          while(len > 1 && !s[len-1]) len--;      }      bign operator * (const bign &b) //*      {          bign c;          c.len = len + b.len;          for(int i = 0; i < len; i++)          {              for(int j = 0; j < b.len; j++)              {                  c.s[i+j] += s[i] * b.s[j];              }          }          for(int i = 0; i < c.len; i++)          {              c.s[i+1] += c.s[i]/10;              c.s[i] %= 10;          }          c.clean();          return c;      }      bign operator *= (const bign &b)      {          *this = *this * b;          return *this;      }      bign operator - (const bign &b)      {          bign c;          c.len = 0;          for(int i = 0, g = 0; i < len; i++)          {              int x = s[i] - g;              if(i < b.len) x -= b.s[i];              if(x >= 0) g = 0;              else              {                  g = 1;                  x += 10;              }              c.s[c.len++] = x;          }          c.clean();          return c;      }      bign operator -= (const bign &b)      {          *this = *this - b;          return *this;      }      bign operator / (const bign &b)      {          bign c, f = 0;          for(int i = len-1; i >= 0; i--)          {              f = f*10;              f.s[0] = s[i];              while(f >= b)              {                  f -= b;                  c.s[i]++;              }          }          c.len = len;          c.clean();          return c;      }      bign operator /= (const bign &b)      {          *this  = *this / b;          return *this;      }      bign operator % (const bign &b)      {          bign r = *this / b;          r = *this - r*b;          return r;      }      bign operator %= (const bign &b)      {          *this = *this % b;          return *this;      }      bool operator < (const bign &b)      {          if(len != b.len) return len < b.len;          for(int i = len-1; i >= 0; i--)          {              if(s[i] != b.s[i]) return s[i] < b.s[i];          }          return false;      }      bool operator > (const bign &b)      {          if(len != b.len) return len > b.len;          for(int i = len-1; i >= 0; i--)          {              if(s[i] != b.s[i]) return s[i] > b.s[i];          }          return false;      }      bool operator == (const bign &b)      {          return !(*this > b) && !(*this < b);      }      bool operator != (const bign &b)      {          return !(*this == b);      }      bool operator <= (const bign &b)      {          return *this < b || *this == b;      }      bool operator >= (const bign &b)      {          return *this > b || *this == b;      }      string str() const      {          string res = "";          for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;          return res;      }  };  istream& operator >> (istream &in, bign &x)  {      string s;      in >> s;      x = s.c_str();      return in;  }  ostream& operator << (ostream &out, const bign &x)  {      out << x.str();      return out;  }  bool vis[305];bign f[305];bign F(int x) {    if (x == 1) return 2;    if (x == 2) return 9;    if (x == 3) return 38;    if (vis[x]) return f[x];    vis[x] = 1;     if (x % 2)        return f[x] = (F(x-1) * 4 + 2);    else {//      cout << F(x-1) * 4 << "\n";//      cout << F(x-3) * 4 << "\n";//      cout << F(x-2) * 4 << "\n";//      cout << ( F(x-3) * 4 - F(x-2) ) * 4 << "\n";        return f[x] = ( F(x-1) * 4 + F(x-2) * 4 - F(x-3) * 16 - 21);     } }int main()  {      freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    int n; cin >> n;    n /= 3;    cout << F(n) << "\n";    return 0;  }

C
题意：给定一个h×w的矩阵，需要用2×2的矩阵去覆盖它，覆盖的矩阵可以重叠但至少需要露出一半的面积。询问最大能放多少个2×2矩阵并且给出一种方案
解法：摆出来的结果被我们成为蛇形矩阵，感受一下吧

#include<bits/stdc++.h>#define N 500010using namespace std;int n,m,i,j,tot,x[N],y[N];int main(){    freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    cin>>n>>m;    if (n<2||m<2) return puts("0"),0;    if (m&1){        for (i=1;i<n;i+=2){            ++tot;            x[tot]=i;            y[tot]=m-1;        }        --m;    }    for (i=n-1;i>1;i--) for (j=1;j<m;j+=2)        ++tot,x[tot]=i,y[tot]=j;    for (i=m-1;i>=1;i--) ++tot,x[tot]=1,y[tot]=i;    cout<<tot<<"\n";    for (i=1;i<=tot;i++) cout<<x[i]<<" "<<y[i]<<"\n";    return 0;}

D
题意：给定一个有4种字符WESN构成的字符串，并且有合法拆分SW,NW,SE,NE或者拆成单个字符串，求合法方案数
解法：签到dp

#include<stdio.h>#include<cstring>#define N 100005#define mod 1000000007char S[N];int n,dp[N];int main(){    freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    scanf("%s",S+1);    n=strlen(S+1);    dp[0]=1;    for (int i=1;i<=n;i++)    {        dp[i]=(dp[i]+dp[i-1]) % mod;        if ((S[i]=='W' || S[i]=='E') && (S[i-1]=='S' || S[i-1]=='N')) dp[i]=(dp[i]+dp[i-2]) % mod;    }    printf("%d",dp[n]);}

EFGHJ
留坑待填

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