2018 Multi-University Training Contest 8 Solution

A - Character Encoding

题意：用m个$0-n-1$的数去构成k，求方案数

思路：当没有0-n-1这个条件是答案为C(k+m-1, m-1),减去有大于的关于n的情况，当有i个n时的种类为C(k+m-1-i*n,m-1)个，利用容斥定理可得

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const int MOD = ;
 const int maxn = 1e6 + ;

 ll fac[maxn];
 ll inv[maxn];
 ll invfac[maxn];

 void Init()
 {
     fac[] = inv[] = invfac[] = ;
     fac[] = inv[] = invfac[] = ;
     for(int i = ; i < maxn; ++i)
     {
         fac[i] = fac[i - ] * i % MOD;
         inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
         invfac[i] = invfac[i - ] * inv[i] % MOD;
     }
 }

 ll calc(ll n, ll m)
 {
     if(m > n || m <  || n < ) return ;
     return fac[n] * invfac[m] % MOD * invfac[n - m] % MOD;
 }

 int n, m , k;

 int main()
 {
     Init();
     int t;
     scanf("%d", &t);
     while(t--)
     {
         scanf("%d %d %d", &n, &m, &k);
         ll ans = ;
         for(int i = ; i <= m; ++i)
         {
             ll tmp = k - 1ll * i * n + m - ;
             if(tmp < ) break;
             if(i & ) ans = (ans - calc(m, i) * calc(tmp, m - ) % MOD + MOD) % MOD;
             else ans = (ans + calc(m, i) * calc(tmp, m - ) % MOD) % MOD;
         //    cout << i << " " << m << " " << calc(m, i);
         //    cout << tmp << " " << m - 1 << " " << calc(tmp, m - 1) << endl;
         }
         printf("%lld\n", ans);
     }
     return ;
 }

B - Pizza Hub

题意：给出一个三角形，以及一个矩形的宽度，求一个最小的高度使得矩形能够覆盖三角形

思路：显然一定有一个点在矩形的顶点，枚举计算即可

 #include<bits/stdc++.h>

 using namespace std;

 const int INF = 0x3f3f3f3f;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 const int maxn = 1e2;

 int sgn(double x)
 {
     if(fabs(x) < eps) return ;
     else return x >  ?  : -;
 }

 struct Point{
     double x, y;
     Point(){}
     Point(double _x, double _y)
     {
         x = _x;
         y = _y;
     }

     void input()
     {
         scanf("%lf %lf", &x, &y);
     }

     Point operator - (const Point &b) const
     {
         return Point(x - b.x, y - b.y);
     }

     double operator ^ (const Point &b) const
     {
         return x * b.y - y * b.x;
     }

     double distance(Point p)
     {
         return hypot(x - p.x, y - p.y);
     }

     double len()
     {
         return hypot(x, y);
     }

     double operator * (const Point &b) const
     {
         return x * b.x + y * b.y;
     }
 }P[maxn];

 int w;
 double ans;
 double area;

 void calc(Point a, Point b)//low high
 {
     if(sgn(a * b) < ) return ;
     if(sgn(a * a - w * w) <= )
     {
         if(sgn(a ^ b) < ) return ;
         if(sgn((a * b) * (a * b) - w * w * (a * a)) > ) return ;
         ans = min(ans, (a ^ b) / sqrt(a * a));
     }
     else
     {
         double h = sqrt(a * a - w * w);
         double src1 = atan(h / w);
         if(sgn(a ^ b) >= )
         {
             double src2 = acos((a * b) / (sqrt(a * a) * sqrt(b * b)));
             if(sgn(src1 + src2 - PI / ) > ) return ;
             double len1 = sqrt(b * b) * cos(src1 + src2);
             if(sgn(len1 - w) > ) return ;
             ans = min(ans, max(h, sqrt(b * b) * sin(src1 + src2)));
         }
         else
         {
             double src2 = acos((a * b) / (sqrt(a * a) * sqrt(b * b)));
             if(sgn(src1 - src2) < ) return ;
             double len1 = sqrt(b * b) * cos(src1 - src2);
             if(sgn(len1 - w) > ) return ;
             ans = min(ans, h);
         }
     }
 }

 int main()
 {
     int t;
     scanf("%d", &t);
     while(t--)
     {
         for(int i = ; i <= ; ++i) P[i].input();
     //    double a = p[1].distance(p[2]);
     //    double b = p[2].distance(p[3]);
     //    double c = p[3].distance(p[1]);
     //    double p = (a + b + c) / 2.0;
     //    area = sqrt(p * (p - a) * (p - b) * (p - c));
         ans = INF * 1.0;
         scanf("%d", &w);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         for(int i = ; i <= ; ++i) P[i].y *= -1.0;
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         calc(P[] - P[], P[] - P[]);
         if(ans >= INF * 1.0) puts("impossible");
         else printf("%.10f\n", ans);
     }
     return ;
 }

C - City Development

留坑。

D - Parentheses Matrix

题意：构造一个矩阵，每一行和每一列都会构成一个括号序列，求合法括号序列尽量多

思路：

分类讨论：

n, m  都是奇数  随便构造

n, m 有一个奇数 那么 答案是那个奇数

比如 1 4

()()

两个都是偶数  如果有一个为2

比如 2 4

((((

))))

如果两个都大于四

比如六 六

((((((

()()()

(()())

()()()

(()())

)))))

像这样构造 答案是 n + m - 4

如果 n, m 中有个4

比如  4 4

(())

()()

(())

()()

 #include<bits/stdc++.h>

 using namespace std;

 const int maxn =  + ;

 int n, m;
 char str[maxn][maxn];

 int main()
 {
     int t;
     scanf("%d", &t);
     while(t--)
     {
         scanf("%d %d", &n, &m);
 //        printf("%d %d\n", n, m);
         if(n %  ==  && m %  == )
         {
             for(int i = ; i <= n; ++i)
             {
                 for(int j = ; j <= m; ++j)
                 {
                     printf("(");
                 }
                 printf("\n");
             }
         }
         else if(n %  == )
         {
             for(int i = ; i <= n; ++i)
             {
                 for(int j = ; j <= m; ++j)
                 {
                     if(j & ) printf("(");
                     else printf(")");
                 }
                 printf("\n");
             }
         }
         else if(m %  == )
         {
             for(int i = ; i <= n; ++i)
             {
                 for(int j = ; j <= m; ++j)
                 {
                     if(i & ) printf("(");
                     else printf(")");
                 }
                 printf("\n");
             }
         }
         else if(n == )
         {
             for(int i = ; i <= n; ++i)
             {
                 for(int j = ; j <= m; ++j)
                 {
                     if(i & ) printf("(");
                     else printf(")");
                 }
                 printf("\n");
             }
         }
         else if(m == )
         {
             for(int i = ; i <= n; ++i)
             {
                 for(int j = ; j <= m; ++j)
                 {
                     if(j & ) printf("(");
                     else printf(")");
                 }
                 printf("\n");
             }
         }
         else if(n == )
         {
             for(int i = ; i <= m; ++i) printf("(");
             printf("\n");
             for(int i = ; i <= m; ++i)
             {
                 if(i & ) printf("(");
                 else printf(")");
             }
             printf("\n");
             for(int i = ; i <= m; ++i)
             {
                 if(i & ) printf(")");
                 else printf("(");
             }
             printf("\n");
             for(int i = ; i <= m; ++i) printf(")");
             printf("\n");
         }
         else if(m == )
         {
             for(int i = ; i <= n; ++i) str[i][] = '(';
             for(int i = ; i <= n; ++i)
             {
                 if(i & ) str[i][] = '(';
                 else str[i][] = ')';
             }
             for(int i = ; i <= n; ++i)
             {
                 if(i & ) str[i][] = ')';
                 else str[i][] = '(';
             }
             for(int i = ; i <= n; ++i) str[i][] = ')';
             for(int i = ; i <= n; ++i)
             {
                 for(int j = ; j <= m; ++j)
                 {
                     printf("%c", str[i][j]);
                 }
                 printf("\n");
             }
         }
         else
         {
             for(int i = ; i <= m; ++i) str[][i] = '(';
             for(int i = ; i <= n; ++i)
             {
                 for(int j = ; j <= m; ++j)
                 {
                     if(i & )
                     {
                         if(j == ) str[i][j] = '(';
                         else if(j == m) str[i][j] = ')';
                         else if(j & ) str[i][j] = ')';
                         else str[i][j] = '(';
                     }
                     else
                     {
                         if(j & ) str[i][j] = '(';
                         else str[i][j] = ')';
                     }
                 }
             }
             for(int i = ; i <= m; ++i) str[n][i] = ')';
             for(int i = ; i <= n; ++i)
             {
                 for(int j = ; j <= m; ++j)
                 {
                     printf("%c", str[i][j]);
                 }
                 printf("\n");
             }
         }
     }
     return ;
 }

E - Magic Square

按题意模拟即可。

 #include <bits/stdc++.h>
 using namespace std;

 #define pii pair <int, int>
 int t, n;
 char G[][];

 void C(pii a)
 {
     int x = a.first, y = a.second;
     char tmp[][];
     tmp[][] = G[x + ][y];
     tmp[][] = G[x][y];
     tmp[][] = G[x + ][y + ];
     tmp[][] = G[x][y + ];
     for (int i = ; i <= ; ++i)
         for (int j = ; j <= ; ++j)
             G[x + i - ][y + j - ] = tmp[i][j];
 }

 void R(pii a)
 {
     int x = a.first, y = a.second;
     char tmp[][];
     tmp[][] = G[x][y + ];
     tmp[][] = G[x + ][y + ];
     tmp[][] = G[x][y];
     tmp[][] = G[x + ][y];
     for (int i = ; i <= ; ++i)
         for (int j = ; j <= ; ++j)
             G[x + i - ][y + j - ] = tmp[i][j];
 }

 pii pos[] =
 {
     pii(, ),
     pii(, ),
     pii(, ),
     pii(, ),
 };

 int main()
 {
     scanf("%d", &t);
     while (t--)
     {
         scanf("%d", &n);
         for (int i = ; i <= ; ++i) scanf("%s", G[i] + );
         char op; int x;
         for (int i = ; i <= n; ++i)
         {
             scanf("%d %c", &x, &op);
             if (op == 'C') C(pos[x - ]);
             else R(pos[x - ]);
         }
         for (int i = ; i <= ; ++i) printf("%s\n", G[i] + );
     }
     return ;
 }

F - Boolean 3-Array

留坑。

G - Card Game

留坑。

H - K-Similar Strings

留坑。

I - Make ZYB Happy

留坑。

J - Taotao Picks Apples

题意：每次可以改变一个值，求改变之后以第一个数开头的最长上升子序列的长度，改变仅当次有效

思路：考虑线段树，维护一个cnt, 和一个Max  两个区间合并的时候

如果左区间的$Max > 右区间的 Max$

那么右区间没有贡献

否则递归查找贡献

 #include <bits/stdc++.h>
 using namespace std;

 #define N 100010
 int t;
 int n, m;
 int arr[N];

 struct SEG
 {
     int cnt[N << ], Max[N << ];
     void build(int id, int l, int r)
     {
         cnt[id] = Max[id] = ;
         if (l == r) return;
         int mid = (l + r) >> ;
         build(id << , l, mid);
         build(id <<  | , mid + , r);
     }
     int query(int id, int l, int r, int val)
     {
         if (l == r) return Max[id] > val;
         int mid = (l + r) >> ;
         if (Max[id] <= val) return ;
         if (Max[id << ] <= val) return query(id <<  | , mid + , r, val);
         else return cnt[id] - cnt[id << ] + query(id << , l, mid, val);
     }
     void update(int id, int l, int r, int pos, int val)
     {
         if (l == r)
         {
             cnt[id] = ;
             Max[id] = val;
             return;
         }
         int mid = (l + r) >> ;
         if (pos <= mid) update(id << , l, mid, pos, val);
         else update(id <<  | , mid + , r, pos, val);
         cnt[id] = cnt[id << ] + query(id <<  | , mid + , r, Max[id << ]);
         Max[id] = max(Max[id << ], Max[id <<  | ]);
     }
 }seg;

 int main()
 {
     scanf("%d", &t);
     while (t--)
     {
         scanf("%d%d", &n, &m);
         for (int i = ; i <= n; ++i) scanf("%d", arr + i);
         seg.build(, , n);
         for (int i = ; i <= n; ++i) seg.update(, , n, i, arr[i]);
         for (int i = , x, v; i <= m; ++i)
         {
             scanf("%d%d", &x, &v);
             seg.update(, , n, x, v);
             printf("%d\n", seg.cnt[]);
             seg.update(, , n, x, arr[x]);
         }
     }
     return ;
 }

K - Pop the Balloons

留坑。

L - From ICPC to ACM

留坑。