Day 1 模拟

1. P1088 火星人

利用STL中的next_permutation();函数求一种排列的下一种排列，循环m次即为答案。(STL大法好~~C++是世界上最好的语言~~逃

 #include <cstdio>
 #include <algorithm>
 using namespace std;
 const int MAXN = ;
 int n, m, s[MAXN];
 int main()
 {
     scanf("%d%d", &n, &m);
     for(int i = ; i <= n - ; ++i)
         scanf("%d", &s[i]);
     for(int i = ; i <= m; ++i)
         next_permutation(s, s + n);
     for(int i = ; i <= n - ; ++i)
         printf("%d ", s[i]);
     return ;
 }

2. P1015 回文数

将高精加推广到n进制即可，不再赘述，代码如下。高精 * 1

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 const int MAXL = ;
 int n, ans;
 string s1, map = "0123456789ABCDEF";
 bool judge(string s1)
 {
     string s2 = s1;
     reverse(s2.begin(), s2.end());
     if(s1 == s2) return true;
     else return false;
 }
 string add(string s1)
 {
     int a[MAXL], b[MAXL], c[MAXL], len = s1.size(), lenc = s1.size();
     memset(a, , sizeof(a));
     memset(b, , sizeof(b));

     memset(c, , sizeof(c));
     string s2 = s1, sum;
     reverse(s2.begin(), s2.end());
     for(int i = ; i <= len; ++i)
     {
         if(isdigit(s1[len - i])) a[i] = s1[len - i] - '';
         else a[i] = s1[len - i] - 'A' + ;
         if(isdigit(s2[len - i])) b[i] = s2[len - i] - '';
         else b[i] = s2[len - i] - 'A' + ;
     }
     for(int i = ; i <= lenc; ++i)
     {
         c[i] += a[i] + b[i];

         if(c[i] >= n)
         {
             c[i] %= n;
             ++c[i + ];
         }
     }
     ++lenc;
     while(c[lenc] ==  && lenc > ) --lenc;
     for(int i = lenc; i >= ; --i)
         sum += map[c[i]];
     return sum;
 }
 int main()
 {
     cin >> n >> s1;
     for(int i = ; i <= ; ++i)
     {
         s1 = add(s1);

         if(judge(s1) == true)
         {
             cout << "STEP=" << i << endl;
             return ;
         }
     }
     cout << "Impossible!" << endl;
     return ;
 } 

 3. P1604 B进制星球

与上题基本同理。高精 * 2

 #include <cstdio>
 #include <cstring>
 using namespace std;
 const int MAXN = ;
 int k, a[MAXN], b[MAXN], c[MAXN];
 char s1[MAXN], s2[MAXN];
 int max(int a, int b)
 {
     if(a >= b) return a;
     return b;
 }
 int main()
 {
     scanf("%d%s%s", &k, s1, s2);
     a[] = strlen(s1);
     b[] = strlen(s2);
     for(int i = ; i <= a[]; ++i)
     {
         if(s1[a[] - i] >= '' && s1[a[] - i] <= '')
             a[i] = s1[a[] - i] - '';
         else if(s1[a[] - i] >= 'A' && s1[a[] - i] <= 'Z')
             a[i] = s1[a[] - i] - 'A' + ;
     }
     for(int i = ; i <= b[]; ++i)
     {
         if(s2[b[] - i] >= '' && s2[b[] - i] <= '')
             b[i] = s2[b[] - i] - '';
         else if(s2[b[] - i] >= 'A' && s2[b[] - i] <= 'Z')
             b[i] = s2[b[] - i] - 'A' + ;
     }
     int len = max(a[], b[]);
     for(int i = ; i <= len; ++i)
     {
         c[i] += a[i] + b[i];
         if(c[i] >= k)
         {
             c[i] %= k;
             ++c[i + ];
         }
     }
     ++len;
     while(c[len] ==  && len > ) --len;
     for(int i = len; i >= ; --i)
     {
         if(c[i] <= )
             printf("%c", char(c[i] + ''));
         else if(c[i] >= )
             printf("%c", char(c[i] -  + 'A'));
     }
     return ;
 }

4. P1402 乒乓球

字符串模拟，码农题。

 #include <iostream>
 using namespace std;
 const int MAXN = ;
 int now = , win[MAXN], w = , l = ;
 char c;
 int main()
 {
     for(int i = ; cin >> c && c != 'E'; ++i)
     {
         if(c == 'W') win[i] = ;
         else if(c == 'L') win[i] = ;
     }
     for(int i = ; ; ++i)
     {
         if(win[i] == )
         {
             cout << w << ":" << l << endl;
             w = ;
             l = ;
             break;
         }
         else if(win[i] == ) ++w;
         else if(win[i] == ) ++l;
         if((w - l >=  || l - w >= ) && (w >=  || l >= ))
         {
             cout << w << ":" << l << endl;
             w = ;
             l = ;
         }
     }
     cout << endl;
     for(int i = ; ; ++i)
     {
         if(win[i] == )
         {
             cout << w << ":" << l << endl;
             w = ;
             l = ;
             break;
         }
         else if(win[i] == ) ++w;
         else if(win[i] == ) ++l;
         if((w - l >=  || l - w >= ) && (w >=  || l >= ))
         {
             cout << w << ":" << l << endl;
             w = ;
             l = ;
         }
     }
     return ;
 }

5. P1255 数楼梯

递推，第k阶的方法=第k-1阶+第k-2阶，其中k的数量比较大，需用高精。高精 * 3

 #include <cstdio>
 using namespace std;
 const int MAXL = , MAXN = ;
 int n, f[MAXN][MAXL];
 int max(int a, int b)
 {
     if(a >= b) return a;
     return b;
 }
 void add(int k)
 {
     int len = max(f[k - ][], f[k - ][]);
     for(int i = ; i <= len; ++i)
     {
         f[k][i] += f[k - ][i] + f[k - ][i];
         if(f[k][i] >= )
         {
             f[k][i] %= ;
             ++f[k][i + ];
         }
     }
     ++len;
     while(f[k][len] ==  && len > ) --len;
     f[k][] = len;
     return ;
 }
 int main()
 {
     f[][] = ;
     f[][] = ;
     f[][] = ;
     f[][] = ;
     f[][] = ;
     f[][] = ;
     scanf("%d", &n);
     if(n ==  || n ==  || n == )
     {
         printf("%d\n", f[n][]);
         return ;
     }
     for(int i = ; i <= n; ++i)
         add(i);
     for(int i = f[n][]; i >= ; --i)
         printf("%d", f[n][i]);
     printf("\n");
     return ;
 } 

6. P1192 台阶问题

和上一题差不多，只是每次上迈的阶数从2变为n，另外本题不用高精，答案%100003

 #include <cstdio>
 using namespace std;
 const int MAXN = ;
 int n, k;
 long long f[MAXN];
 int min(int a, int b)
 {
     if(a <= b) return a;
     return b;
 }
 int main()
 {
     scanf("%d%d", &n, &k);
     f[] = ;
     f[] = ;
     for(int i = ; i <= n; ++i)
         for(int j = min(i, k); j >= ; --j)
         {
             f[i] += f[i - j];
             if(f[i] >= ) f[i] %= ;
         }
     printf("%lld\n", f[n]);
     return ;
 } 

By ZRQ