A - Absolute
留坑。
B - Counting Permutations
留坑。
C - Cover
留坑。
D - Game
puts("Yes")
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int n; 6 7 int main() 8 { 9 while (scanf("%d", &n) != EOF) 10 { 11 puts("Yes"); 12 } 13 return 0; 14 }
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$sum_{i = l} ^ {i = r} lfloor frac{a_i}{b_i} floor$
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
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define N 100010 6 #define ll long long 7 8 ll arr[N]; 9 10 struct node 11 { 12 int l, r; 13 ll Min, lazy, sum, v; 14 inline node() {} 15 inline node(int _l, int _r) 16 { 17 l = _l; r = _r; 18 Min = 0, lazy = 0, sum = 0, v = 0; 19 } 20 }tree[N << 2]; 21 22 inline void pushup(int id) 23 { 24 tree[id].Min = min(tree[id << 1].Min, tree[id << 1 | 1].Min); 25 tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum; 26 } 27 28 inline void pushdown(int id) 29 { 30 if(tree[id].l >= tree[id].r) return; 31 if(tree[id].lazy) 32 { 33 tree[id << 1].lazy += tree[id].lazy; 34 tree[id << 1 | 1].lazy += tree[id].lazy; 35 tree[id << 1].Min -= tree[id].lazy; 36 tree[id << 1 | 1].Min -= tree[id].lazy; 37 tree[id].lazy = 0; 38 } 39 } 40 41 inline void build(int id, int l, int r) 42 { 43 tree[id] = node(l, r); 44 if (l == r) 45 { 46 tree[id].v = arr[l]; 47 tree[id].Min = arr[l]; 48 return; 49 } 50 int mid = (l + r) >> 1; 51 build(id << 1, l, mid); 52 build(id << 1 | 1, mid + 1, r); 53 pushup(id); 54 } 55 56 inline void update(int id, int l, int r) 57 { 58 if (tree[id].l == l && tree[id].r == r && tree[id].Min > 1) 59 { 60 tree[id].lazy++; 61 tree[id].Min--; 62 return ; 63 } 64 if(tree[id].l == tree[id].r && tree[id].Min == 1) 65 { 66 tree[id].Min = tree[id].v; 67 tree[id].sum++; 68 return ; 69 } 70 pushdown(id); 71 int mid = (tree[id].l + tree[id].r) >> 1; 72 if (r <= mid) update(id << 1, l, r); 73 else if (l > mid) update(id << 1 | 1, l, r); 74 else 75 { 76 update(id << 1, l, mid); 77 update(id << 1 | 1, mid + 1, r); 78 } 79 pushup(id); 80 } 81 82 ll anssum; 83 84 inline void query(int id, int l, int r) 85 { 86 if (tree[id].l >= l && tree[id].r <= r) 87 { 88 anssum += tree[id].sum; 89 return; 90 } 91 pushdown(id); 92 int mid = (tree[id].l + tree[id].r) >> 1; 93 if (l <= mid) query(id << 1, l, r); 94 if (r > mid) query(id << 1 | 1, l, r); 95 pushup(id); 96 } 97 98 int n, m; 99 char str[100]; 100 int l, r; 101 102 int main() 103 { 104 while(~scanf("%d %d", &n, &m)) 105 { 106 for(int i = 1; i <= n; ++i) scanf("%lld", arr + i); 107 build(1, 1, n); 108 for(int i = 1; i <= m; ++i) 109 { 110 scanf("%s", str); 111 if(str[0] == 'a') 112 { 113 scanf("%d %d", &l, &r); 114 update(1, l, r); 115 } 116 else 117 { 118 scanf("%d %d", &l, &r); 119 anssum = 0; 120 query(1, l, r); 121 printf("%lld ", anssum); 122 } 123 } 124 } 125 return 0; 126 }
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
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define ll long long 6 #define N 100010 7 8 int n, m; ll x, y; 9 int arr[N], brr[N]; 10 int a[N]; 11 12 inline void Init() 13 { 14 for (int i = 1; i <= n; ++i) brr[i] = arr[i]; 15 sort(brr + 1, brr + 1 + n); 16 m = unique(brr + 1, brr + 1 + n) - brr - 1; 17 } 18 19 inline int Get(int x) 20 { 21 return lower_bound(brr + 1, brr + 1 + m, x) - brr; 22 } 23 24 inline int lowbit(int x) 25 { 26 return x & (-x); 27 } 28 29 inline void update(int x, int val) 30 { 31 for (int i = x; i <= n; i += lowbit(i)) 32 a[i] += val; 33 } 34 35 inline int query(int x) 36 { 37 int res = 0; 38 for (int i = x; i > 0; i -= lowbit(i)) 39 res += a[i]; 40 return res; 41 } 42 43 int main() 44 { 45 while (scanf("%d%lld%lld", &n, &x, &y) != EOF) 46 { 47 for (int i = 1; i <= n; ++i) scanf("%d", arr + i); 48 Init(); memset(a, 0, sizeof a); 49 for (int i = 1; i <= n; ++i) arr[i] = Get(arr[i]); 50 ll ans = 0; 51 for (int i = 1; i <= n; ++i) 52 { 53 // ans += i - 1 - query(arr[i] - 1); 54 // printf("%d %d ", arr[i], query(arr[i] - 1)); 55 update(arr[i], 1); 56 ans += i - query(arr[i]); 57 // cout << arr[i] << " " << query(arr[i]) << endl; 58 } 59 printf("%lld ", min(ans * x, ans * y)); 60 } 61 return 0; 62 }