A. Hard to prepare
题意：
思路：
$\sum_{i = 0}^{i = n - 1} C_n^i \cdot t^{n - 1} * (-1)^i$ (n 是奇数)
$\sum_{i = 0}^{i = n - 1} C_n^i \cdot t^{n - 1} * (-1)^i + t$ (n 是偶数)
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 1000010 6 7 const ll MOD = (ll)1e9 + 7; 8 9 int t, n, k; 10 ll inv[N]; 11 ll Bit[N]; 12 ll Bitt[N]; 13 14 inline void Init() 15 { 16 inv[1] = 1; 17 for (int i = 2; i < N; ++i) inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD; 18 Bit[0] = 1; 19 for (int i = 1; i < N; ++i) Bit[i] = (Bit[i - 1] * 2) % MOD; 20 } 21 22 inline void init() 23 { 24 Bitt[0] = 1; ll tmp = Bit[k]; 25 for (int i = 1; i <= n; ++i) Bitt[i] = (Bitt[i - 1] * tmp) % MOD; 26 } 27 28 inline void Run() 29 { 30 scanf("%d", &t); Init(); 31 while (t--) 32 { 33 scanf("%d%d", &n, &k); 34 if (k == 1) 35 { 36 puts("2"); 37 continue; 38 } 39 else if (n == 1) 40 { 41 printf("%lld\n",Bit[k]); 42 continue; 43 } 44 init(); 45 ll res = 0; 46 ll C = 1; 47 for (int i = 0; i < n; ++i) 48 { 49 res = (res + (C * Bitt[n - i] % MOD * ((i & 1) ? -1 : 1) + MOD) % MOD) % MOD; 50 C = C * (n - i) % MOD * inv[i + 1] % MOD; 51 } 52 res = (res + (Bit[k] * ((n & 1) ? 0 : 1)) % MOD + MOD) % MOD; 53 printf("%lld\n", res); 54 } 55 } 56 57 int main() 58 { 59 #ifdef LOCAL 60 freopen("Test.in", "r", stdin); 61 #endif 62 63 Run(); 64 return 0; 65 }
B. BE, GE or NE
题意：每一轮有三种操作， 加上a 减去b 或者 取负 当且仅当 a, b, c 不为0时，对应的操作有效，给出一个上界和一个下界 大于等于上界就是 Good Ending 小于等于下界 就是 Bad Ending 否则就是 Normal Ending 两个人轮流操作，第一个人想要Good Ending 第二个人想要 Bad Ending 两个人操作最优，求最后的结局
思路：dp[i][j] 表示 第几轮 数字是多少的时候 ，记忆化爆搜 因为数字在$[-100, 100]$
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const int MOD = (int)1e9 + 7; 8 const int INF = 0x3f3f3f3f; 9 const ll INFLL = 0x3f3f3f3f3f3f3f3f; 10 const int maxn = (int)1e3 + 10; 11 12 struct node { 13 int a, b, c; 14 inline node() {} 15 inline node(int a, int b, int c) :a(a), b(b), c(c) {} 16 }arr[maxn]; 17 18 int n, m, l, k; 19 int type[maxn][maxn];//1 good 0 normal -1 bad 20 int vis[maxn][maxn]; 21 22 inline void Init() 23 { 24 memset(vis, 0, sizeof vis); 25 memset(type, 0, sizeof type); 26 } 27 28 inline int DFS(int idx, int state) 29 { 30 if (idx > n) 31 { 32 if (state <= l) return -1; 33 else if (state >= k) return 1; 34 else return 0; 35 } 36 if (vis[idx][state]) return type[idx][state]; 37 vis[idx][state] = 1; 38 int res = 0; 39 //a 40 int A = 2, B = 2, C = 2; 41 if (arr[idx].a) 42 { 43 A = DFS(idx + 1, min(100, state + arr[idx].a)); 44 } 45 //b 46 if (arr[idx].b) 47 { 48 B = DFS(idx + 1, max(-100, state - arr[idx].b)); 49 } 50 //c 51 if (arr[idx].c) 52 { 53 C = DFS(idx + 1, max(-100, min(100, state * -arr[idx].c))); 54 } 55 if ((A == 1 || A == 2)&& (B == 1 || B == 2) && (C == 1 || C == 2) && (idx & 1) == 0) 56 { 57 type[idx][state] = 1; 58 return 1; 59 } 60 else if ((A == -1 || A == 2) && (B == -1 || B == 2) && (C == -1 || C == 2) && (idx & 1) == 1) 61 { 62 type[idx][state] = -1; 63 return -1; 64 } 65 else if ((A == 1 || B == 1 || C == 1) && (idx & 1) == 1) 66 { 67 type[idx][state] = 1; 68 return 1; 69 } 70 else if ((A == -1 || B == -1 || C == -1) && (idx & 1) == 0) 71 { 72 type[idx][state] = -1; 73 return -1; 74 } 75 else 76 { 77 type[idx][state] = 0; 78 return 0; 79 } 80 } 81 82 inline void RUN() 83 { 84 while (~scanf("%d %d %d %d", &n, &m, &k, &l)) 85 { 86 Init(); 87 for (int i = 1; i <= n; ++i) 88 { 89 scanf("%d %d %d", &arr[i].a, &arr[i].b, &arr[i].c); 90 } 91 int ans = DFS(1, m); 92 if (ans == 1) 93 { 94 puts("Good Ending"); 95 } 96 else if (ans == -1) 97 { 98 puts("Bad Ending"); 99 } 100 else 101 { 102 puts("Normal Ending"); 103 } 104 } 105 } 106 107 int main() 108 { 109 #ifdef LOCAL_JUDGE 110 freopen("Text.txt", "r", stdin); 111 #endif // LOCAL_JUDGE 112 113 RUN(); 114 115 #ifdef LOCAL_JUDGE 116 fclose(stdin); 117 #endif // LOCAL_JUDGE 118 }
C. Cacti Lottery
留坑。
D. Easy Math
留坑。
E. End Fantasy VIX
留坑。
F. Features Track
水。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 #define ll long long 6 7 typedef pair <ll, ll> pii; 8 9 int t, n, tot; 10 ll ans, x, y; 11 map <pii, pii> mp; 12 13 inline void Run() 14 { 15 scanf("%d", &t); 16 while (t--) 17 { 18 scanf("%d", &n); mp.clear(); ans = 0; 19 for (int i = 1; i <= n; ++i) 20 { 21 scanf("%d", &tot); 22 for (int j = 1; j <= tot; ++j) 23 { 24 scanf("%lld%lld", &x, &y); 25 if (mp[pii(x, y)].second == i - 1) ++mp[pii(x, y)].first; 26 else if (mp[pii(x, y)].second == i) continue; 27 else mp[pii(x, y)].first = 1; 28 ans = max(ans, mp[pii(x, y)].first); 29 mp[pii(x, y)].second = i; 30 } 31 } 32 printf("%lld\n", ans); 33 } 34 } 35 36 int main() 37 { 38 #ifdef LOCAL 39 freopen("Test.in", "r", stdin); 40 #endif 41 42 Run(); 43 return 0; 44 }
G. Trace
题意：每次给出一个点，然后就会形成两条线，如果后面的矩形覆盖了前面的边，那么这条边就消失了， 最后求剩下的边是多少
思路：分别处理x轴，y轴，然后排序，然后扫过去，每次加上自己的边长以及减去标号比自己小的并且长度比自己高的个数乘自己的边长
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 50010 5 #define ll long long 6 7 int n; 8 9 struct node 10 { 11 int l, r; 12 int lazy, sum; 13 inline node() {} 14 inline node(int _l, int _r) 15 { 16 l = _l, r = _r; 17 lazy = -1; 18 sum = 0; 19 } 20 }tree[N << 2]; 21 22 inline void pushup(int id) 23 { 24 tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum; 25 } 26 27 inline void pushdown(int id) 28 { 29 if (tree[id].l >= tree[id].r) return; 30 if (~tree[id].lazy) 31 { 32 int lazy = tree[id].lazy; tree[id].lazy = -1; 33 tree[id << 1].lazy = tree[id << 1 | 1].lazy = lazy; 34 tree[id << 1].sum = tree[id << 1 | 1].sum = 0; 35 } 36 } 37 38 inline void build(int id, int l, int r) 39 { 40 tree[id] = node(l, r); 41 if (l == r) return; 42 int mid = (l + r) >> 1; 43 build(id << 1, l, mid); 44 build(id << 1 | 1, mid + 1, r); 45 } 46 47 inline void update(int id, int l, int r, int val) 48 { 49 if (tree[id].l >= l && tree[id].r <= r) 50 { 51 tree[id].sum = val; 52 tree[id].lazy = val; 53 return; 54 } 55 pushdown(id); 56 int mid = (tree[id].l + tree[id].r) >> 1; 57 if (l <= mid) update(id << 1, l, r, val); 58 if (r > mid) update(id << 1 | 1, l, r, val); 59 pushup(id); 60 } 61 62 inline int query(int id, int l, int r) 63 { 64 if (tree[id].l >= l && tree[id].r <= r) return tree[id].sum; 65 pushdown(id); 66 int mid = (tree[id].l + tree[id].r) >> 1; 67 int res = 0; 68 if (l <= mid) res += query(id << 1, l, r); 69 if (r > mid) res += query(id << 1 | 1, l, r); 70 return res; 71 } 72 73 struct DT 74 { 75 int pos; 76 ll x, y; 77 inline void scan(int _pos) 78 { 79 pos = _pos; 80 scanf("%lld%lld", &x, &y); 81 } 82 }arr[N]; 83 84 inline bool cmp1(DT a, DT b) 85 { 86 return a.x < b.x; 87 } 88 89 inline bool cmp2(DT a, DT b) 90 { 91 return a.y < b.y; 92 } 93 94 inline void Run() 95 { 96 while (scanf("%d", &n) != EOF) 97 { 98 for (int i = 1; i <= n; ++i) arr[i].scan(i); 99 build(1, 1, n); 100 sort(arr + 1, arr + 1 + n, cmp1); 101 ll res = 0; 102 for (int i = 1; i <= n; ++i) 103 { 104 res += arr[i].y; 105 int pos = query(1, 1, arr[i].pos); 106 res -= arr[i].y * pos; 107 update(1, 1, arr[i].pos, 0); 108 update(1, arr[i].pos, arr[i].pos, 1); 109 } 110 sort(arr + 1, arr + 1 + n, cmp2); 111 update(1, 1, n, 0); 112 for (int i = 1; i <= n; ++i) 113 { 114 res += arr[i].x; 115 int pos = query(1, 1, arr[i].pos); 116 res -= arr[i].x * pos; 117 update(1, 1, arr[i].pos, 0); 118 update(1, arr[i].pos, arr[i].pos, 1); 119 } 120 printf("%lld\n", res); 121 } 122 } 123 124 int main() 125 { 126 #ifdef LOCAL 127 freopen("Test.in", "r", stdin); 128 #endif 129 130 Run(); 131 return 0; 132 }
H. Ryuji doesn't want to study
题意：两个操作，第一种是查询$[L, R]$ 区间内 $a[L] * len + a[L + 1] * (len - 1) + ... + a[R] * 1$
第二种是改变一个数
思路：线段树，记录两个值，一个是sum,另一个是 $a[L] * len + a[L + 1] * (len - 1) + ... + a[R] * 1$
考虑合并的时候 显然两个区间合并，相当于左区间的长度增加了右区间的长度，那么只需要多加上左区间的sum * 右区间长度
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 #define ll long long 6 7 int n, q; 8 ll arr[N]; 9 10 struct node 11 { 12 int l, r; 13 ll sum1, sum2; 14 inline node() {} 15 inline node(int l, int r, ll sum1, ll sum2) : l(l), r(r), sum1(sum1), sum2(sum2) {} 16 }tree[N << 2]; 17 18 inline void pushup(int id) 19 { 20 tree[id].sum1 = tree[id << 1].sum1 + tree[id << 1 | 1].sum1; 21 tree[id].sum2 = tree[id << 1 | 1].sum2 + tree[id << 1].sum2 + tree[id << 1].sum1 * (tree[id << 1 | 1].r - tree[id << 1 | 1].l + 1); 22 } 23 24 inline void build(int id, int l, int r) 25 { 26 tree[id] = node(l, r, 0, 0); 27 if (l == r) 28 { 29 tree[id].sum1 = arr[tree[id].l]; 30 tree[id].sum2 = arr[tree[id].l]; 31 return; 32 } 33 int mid = (l + r) >> 1; 34 build(id << 1, l, mid); 35 build(id << 1 | 1, mid + 1, r); 36 pushup(id); 37 } 38 39 inline void update(int id, int pos, ll val) 40 { 41 if (tree[id].l == tree[id].r) 42 { 43 tree[id].sum1 = val; 44 tree[id].sum2 = val; 45 return; 46 } 47 int mid = (tree[id].l + tree[id].r) >> 1; 48 if (pos <= mid) update(id << 1, pos, val); 49 else update(id << 1 | 1, pos, val); 50 pushup(id); 51 } 52 53 ll anssum; 54 55 inline void query(int id, int l, int r) 56 { 57 if (tree[id].l >= l && tree[id].r <= r) 58 { 59 anssum += tree[id].sum2 + tree[id].sum1 * (r - tree[id].r); 60 return; 61 } 62 int mid = (tree[id].l + tree[id].r) >> 1; 63 if (l <= mid) query(id << 1, l, r); 64 if (r > mid) query(id << 1 | 1, l, r); 65 } 66 67 inline void Run() 68 { 69 while (scanf("%d%d", &n, &q) != EOF) 70 { 71 for (int i = 1; i <= n; ++i) scanf("%lld", arr + i); 72 build(1, 1, n); 73 int op, a, b; ll v; 74 for (int i = 1; i <= q; ++i) 75 { 76 scanf("%d", &op); 77 if (op == 1) 78 { 79 scanf("%d%d", &a, &b); 80 anssum = 0; query(1, a, b); 81 printf("%lld\n", anssum); 82 } 83 else 84 { 85 scanf("%d%lld", &a, &v); 86 update(1, a, v); 87 } 88 } 89 } 90 } 91 92 int main() 93 { 94 #ifdef LOCAL 95 freopen("Test.in", "r", stdin); 96 #endif 97 98 Run(); 99 return 0; 100 }
I. Characters with Hash
水。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const int MOD = (int)1e9 + 7; 8 const int INF = 0x3f3f3f3f; 9 const ll INFLL = 0x3f3f3f3f3f3f3f3f; 10 const int maxn = (int)1e6 + 10; 11 12 int n; 13 char s[10]; 14 char str[maxn]; 15 16 int arr[maxn]; 17 18 inline void RUN() 19 { 20 int t; 21 scanf("%d", &t); 22 while (t--) 23 { 24 scanf("%d", &n); 25 scanf("%s", s); 26 scanf("%s", str); 27 int len = strlen(str); 28 for (int i = 0; i < len; ++i) 29 { 30 arr[i] = abs(str[i] - s[0]); 31 } 32 int ans = 2 * len; 33 for (int i = 0; i < len; ++i) 34 { 35 if (arr[i] == 0) 36 { 37 ans -= 2; 38 } 39 else if (arr[i] < 10) 40 { 41 ans -= 1; 42 break; 43 } 44 else 45 { 46 break; 47 } 48 } 49 if (ans == 0) ans = 1; 50 printf("%d\n", ans); 51 } 52 } 53 54 int main() 55 { 56 #ifdef LOCAL_JUDGE 57 freopen("Text.txt", "r", stdin); 58 #endif // LOCAL_JUDGE 59 60 RUN(); 61 62 #ifdef LOCAL_JUDGE 63 fclose(stdin); 64 #endif // LOCAL_JUDGE 65 }
J. Maze Designer
留坑。
K. Morgana Net
留坑。