2018 Multi-University Training Contest 2 Solution

A - Absolute

留坑。

B - Counting Permutations

留坑。

C - Cover

留坑。

D - Game

puts("Yes")

 #include <bits/stdc++.h>

 using namespace std;

 int n;

 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         puts("Yes");
     }
     return ;
 }

E - Hack It

留坑。

F - Matrix

留坑。

G - Naive Operations

题意：给出$b[]$数组，里面是$1-n$ 的全排列，两种操作，一个区间+1，一个是区间求$\sum_{i = l} ^ {i = r} \lfloor \frac{a_i}{b_i} \rfloor$

思路：维护一个Min 表示这个区间内需要的最少的进位，如果有进位，就更新到底，如果没有进位就区间更新

 #include <bits/stdc++.h>

 using namespace std;

 #define N 100010
 #define ll long long

 ll arr[N];

 struct node
 {
     int l, r;
     ll Min, lazy, sum, v;
     inline node() {}
     inline node(int _l, int _r)
     {
         l = _l; r = _r;
         Min = , lazy = , sum = , v = ;
     }
 }tree[N << ];

 inline void pushup(int id)
 {
     tree[id].Min = min(tree[id << ].Min, tree[id <<  | ].Min);
     tree[id].sum = tree[id << ].sum + tree[id <<  | ].sum;
 }

 inline void pushdown(int id)
 {
     if(tree[id].l >= tree[id].r) return;
     if(tree[id].lazy)
     {
         tree[id << ].lazy += tree[id].lazy;
         tree[id <<  | ].lazy += tree[id].lazy;
         tree[id << ].Min -= tree[id].lazy;
         tree[id <<  | ].Min -= tree[id].lazy;
         tree[id].lazy = ;
     }
 }

 inline void build(int id, int l, int r)
 {
     tree[id] = node(l, r);
     if (l == r)
     {
         tree[id].v = arr[l];
         tree[id].Min = arr[l];
         return;
     }
     int mid = (l + r) >> ;
     build(id << , l, mid);
     build(id <<  | , mid + , r);
     pushup(id);
 }

 inline void update(int id, int l, int r)
 {
     if (tree[id].l == l && tree[id].r == r && tree[id].Min > )
     {
         tree[id].lazy++;
         tree[id].Min--;
         return ;
     }
     if(tree[id].l == tree[id].r && tree[id].Min == )
     {
         tree[id].Min = tree[id].v;
         tree[id].sum++;
         return ;
     }
     pushdown(id);
     int mid = (tree[id].l + tree[id].r) >> ;
     if (r <= mid) update(id << , l, r);
     else if (l > mid) update(id <<  | , l, r);
     else
     {
         update(id << , l, mid);
         update(id <<  | , mid + , r);
     }
     pushup(id);
 }

 ll anssum;

 inline void query(int id, int l, int r)
 {
     if (tree[id].l >= l && tree[id].r <= r)
     {
         anssum += tree[id].sum;
         return;
     }
     pushdown(id);
     int mid = (tree[id].l + tree[id].r) >> ;
     if (l <= mid) query(id << , l, r);
     if (r > mid) query(id <<  | , l, r);
     pushup(id);
 }

 int n, m;
 char str[];
 int l, r;

 int main()
 {
     while(~scanf("%d %d", &n, &m))
     {
         for(int i = ; i <= n; ++i) scanf("%lld", arr + i);
         build(, , n);
         for(int i = ; i <= m; ++i)
         {
             scanf("%s", str);
             if(str[] == 'a')
             {
                 scanf("%d %d", &l, &r);
                 update(, l, r);
             }
             else
             {
                 scanf("%d %d", &l, &r);
                 anssum = ;
                 query(, l, r);
                 printf("%lld\n", anssum);
             }
         }
     }
     return ;
 }

H - Odd Shops

留坑。

I - Segment

留坑。

J - Swaps and Inversions

水。逆序对的意义就是每次只能交换相邻两个，最少的交换次数

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long
 #define N 100010

 int n, m; ll x, y;
 int arr[N], brr[N];
 int a[N];

 inline void Init()
 {
     for (int i = ; i <= n; ++i) brr[i] = arr[i];
     sort(brr + , brr +  + n);
     m = unique(brr + , brr +  + n) - brr - ;
 }

 inline int Get(int x)
 {
     return lower_bound(brr + , brr +  + m, x) - brr;
 }

 inline int lowbit(int x)
 {
     return x & (-x);
 }

 inline void update(int x, int val)
 {
     for (int i = x; i <= n; i += lowbit(i))
         a[i] += val;
 }

 inline int query(int x)
 {
     int res = ;
     for (int i = x; i > ; i -= lowbit(i))
         res += a[i];
     return res;
 }

 int main()
 {
     while (scanf("%d%lld%lld", &n, &x, &y) != EOF)
     {
         for (int i = ; i <= n; ++i) scanf("%d", arr + i);
         Init(); memset(a, , sizeof a);
         for (int i = ; i <= n; ++i) arr[i] = Get(arr[i]);
         ll ans = ;
         for (int i = ; i <= n; ++i)
         {
         //    ans += i - 1 - query(arr[i] - 1);
         //    printf("%d %d\n", arr[i], query(arr[i] - 1));
             update(arr[i], );
             ans += i - query(arr[i]);
         //    cout << arr[i] << " " << query(arr[i]) << endl;
         }
         printf("%lld\n", min(ans * x, ans * y));
     }
     return ;
 }