标签：res mes 多重 main for word put oom back
A. Magic Mirror
水。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int t; 5 char s[100]; 6 7 inline bool work() 8 { 9 int len = strlen(s); 10 if (len != 6) return false; 11 if (s[0] != ‘j‘ && s[0] != ‘J‘) return false; 12 if (s[1] != ‘e‘ && s[1] != ‘E‘) return false; 13 if (s[2] != ‘s‘ && s[2] != ‘S‘) return false; 14 if (s[3] != ‘s‘ && s[3] != ‘S‘) return false; 15 if (s[4] != ‘i‘ && s[4] != ‘I‘) return false; 16 if (s[5] != ‘e‘ && s[5] != ‘E‘) return false; 17 return true; 18 } 19 20 inline void Run() 21 { 22 scanf("%d", &t); 23 while (t--) 24 { 25 scanf("%s", s); 26 puts(work() ? "Good guy!" : "Dare you say that again?"); 27 } 28 } 29 30 int main() 31 { 32 #ifdef LOCAL 33 freopen("Test.in", "r", stdin); 34 #endif 35 36 Run(); 37 return 0; 38 }
B. Mathematical Curse
题意：有n个房间,m个诅咒，每个房间有一个数值，刚开始有一个初始值，每次进入一个房间可以选择消除诅咒或者不消除，消除诅咒只能顺序消除，消除诅咒就是拿初始值和房间的数值做运算，求最后最大的数是多少
思路：Max[i][j] 表示 第i个房间 第j个操作的最大值, Min[i][j] 表示第i个房间第j个操作的最小值
因为乘法 负负相乘可能变得很大
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 #define ll long long 6 #define INFLL 0x3f3f3f3f3f3f3f3f 7 8 int t, n, m; 9 int arr[N]; 10 char f[10]; 11 ll Max[N][10], Min[N][10]; 12 ll k, ans; 13 14 inline void Run() 15 { 16 scanf("%d", &t); 17 while (t--) 18 { 19 scanf("%d%d%lld", &n, &m, &k); 20 for (int i = 1; i <= n; ++i) scanf("%d", arr + i); 21 scanf("%s", f + 1); 22 memset(Max, -0x3f, sizeof Max); 23 memset(Min, 0x3f, sizeof Min); 24 Max[0][0] = Min[0][0] = k; ans = -INFLL; 25 for (int i = 1; i <= n; ++i) 26 { 27 Max[i][0] = k, Min[i][0] = k; 28 for (int j = 1; j <= min(m, i); ++j) 29 { 30 Max[i][j] = Max[i - 1][j], Min[i][j] = Min[i - 1][j]; 31 ll a = Max[i - 1][j - 1], c = Min[i - 1][j - 1], b = (ll)arr[i]; 32 if (f[j] == ‘+‘) 33 { 34 a += b, c += b; 35 } 36 else if (f[j] == ‘-‘) 37 { 38 a -= b, c -= b; 39 } 40 else if (f[j] == ‘*‘) 41 { 42 a *= b, c *= b; 43 } 44 else if (f[j] == ‘/‘) 45 { 46 a /= b, c /= b; 47 } 48 if (a < c) swap(a, c); 49 Max[i][j] = max(Max[i][j], a); 50 Min[i][j] = min(Min[i][j], c); 51 //printf("%d %d %lld %lld\n", i, j, Max[i][j], Min[i][j]); 52 } 53 ans = max(ans, Max[i][m]); 54 } 55 printf("%lld\n", ans); 56 } 57 } 58 59 int main() 60 { 61 #ifdef LOCAL 62 freopen("Test.in", "r", stdin); 63 #endif 64 65 Run(); 66 return 0; 67 }
C. Password
留坑。
D. Sequence
留坑。
E. Jiu Yuan Wants to Eat
留坑。
F. Modular Production Line
留坑。
G. Give Candies
题意：有n颗糖，有n个人，按顺序出列，每次随机给那个人一些糖（至少一颗），分完为止，求有多少方案
思路：规律是$2^{n - 1}$ 根据费马小定理 $a^{p - 1} = 1 \pmod p$ 那么 只要 先用 $(n - 1) % (MOD - 1)$ 再快速幂
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #include<map> 7 8 using namespace std; 9 10 typedef long long ll; 11 12 const int MOD = (int)1e9 + 7; 13 const int INF = 0x3f3f3f3f; 14 const int maxn = (int)1e5 + 10; 15 16 inline ll qpow(ll n, ll x) 17 { 18 ll res = 1; 19 while (n) 20 { 21 if (n & 1) res = (res * x) % MOD; 22 x = (x*x) % MOD; 23 n >>= 1; 24 } 25 return res; 26 } 27 28 int t; 29 char str[maxn]; 30 31 inline void RUN() 32 { 33 scanf("%d", &t); 34 while (t--) 35 { 36 scanf("%s", str); 37 ll n = 0; 38 int len = strlen(str); 39 for (int i = 0; i < len; ++i) 40 { 41 n = n * 10 + str[i] - ‘0‘; 42 n %= (MOD - 1); 43 } 44 n = (n - 1 + (MOD - 1)) % (MOD - 1); 45 //cout << n << endl; 46 ll ans = qpow(n, 2); 47 printf("%lld\n", ans); 48 } 49 } 50 51 int main() 52 { 53 #ifdef LOCAL_JUDGE 54 freopen("Text.txt", "r", stdin); 55 #endif // LOCAL_JUDGE 56 57 RUN(); 58 59 #ifdef LOCAL_JUDGE 60 fclose(stdin); 61 #endif // LOCAL_JUDGE 62 63 }
H. String and Times
留坑。
I. Save the Room
水。
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #include<map> 7 8 using namespace std; 9 10 typedef long long ll; 11 12 const int MOD = (int)1e9 + 7; 13 const int INF = 0x3f3f3f3f; 14 const int maxn = (int)1e5 + 10; 15 16 int a, b, c; 17 18 inline void RUN() 19 { 20 while (~scanf("%d %d %d", &a, &b, &c)) 21 { 22 if (a % 2 == 0 || b % 2 == 0 || c % 2 == 0) 23 { 24 puts("Yes"); 25 } 26 else 27 { 28 puts("No"); 29 } 30 } 31 } 32 33 int main() 34 { 35 #ifdef LOCAL_JUDGE 36 freopen("Text.txt", "r", stdin); 37 #endif // LOCAL_JUDGE 38 39 RUN(); 40 41 #ifdef LOCAL_JUDGE 42 fclose(stdin); 43 #endif // LOCAL_JUDGE 44 45 }
J. Participate in E-sports
题意：判断$\frac {(n - 1)(n)}{2}$ 和 $n$ 是不是平方数
思路：打表，二分
K. Transport Ship
题意： 有n种船，每种船有$v[i]$容量，每种船有$2^{c[i]} - 1$ 个，每次询问s，求能把s刚好装下的船的分配方案有多少
思路：多重背包+记录方案数
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 10010 5 #define ll long long 6 #define INF 0x3f3f3f3f 7 8 const ll MOD = (ll)1e9 + 7; 9 10 int Bit[30]; 11 12 inline void Init() 13 { 14 Bit[0] = 1; 15 for (int i = 1; i <= 20; ++i) Bit[i] = (Bit[i - 1] << 1); 16 for (int i = 1; i <= 20; ++i) --Bit[i]; 17 } 18 19 int t, n, q; 20 int v[30], c[30]; 21 ll dp[N]; 22 ll f[N]; 23 24 inline void Run() 25 { 26 Init(); 27 scanf("%d", &t); 28 while (t--) 29 { 30 scanf("%d%d", &n, &q); 31 for (int i = 1; i <= n; ++i) 32 { 33 scanf("%d%d", v + i, c + i); 34 c[i] = Bit[c[i]]; 35 } 36 memset(f, 0, sizeof f); 37 dp[0] = 0; f[0] = 1; 38 int m = 10000; 39 for (int i = 1; i <= m; ++i) dp[i] = -INF; 40 for (int i = 1; i <= n; ++i) 41 { 42 if (v[i] * c[i] >= m) 43 { 44 for (int j = v[i]; j <= m; ++j) 45 { 46 dp[j] = max(dp[j], dp[j - v[i]] + v[i]); 47 if (dp[j] == dp[j - v[i]] + v[i]) 48 f[j] = (f[j] + f[j - v[i]]) % MOD; 49 } 50 } 51 else 52 { 53 int tmp = c[i]; 54 for (int k = 1; k < tmp; tmp -= k, k <<= 1) 55 { 56 for (int j = m; j >= k * v[i]; --j) 57 { 58 int w = k * v[i]; 59 dp[j] = max(dp[j], dp[j - w] + w); 60 if (dp[j] == dp[j - w] + w) 61 f[j] = (f[j] + f[j - w]) % MOD; 62 } 63 } 64 for (int j = m; j >= tmp * v[i]; --j) 65 { 66 int w = tmp * v[i]; 67 dp[j] = max(dp[j], dp[j - w] + w); 68 if (dp[j] == dp[j - w] + w) 69 f[j] = (f[j] + f[j - w]) % MOD; 70 } 71 } 72 } 73 for (int i = 1, id; i <= q; ++i) 74 { 75 scanf("%d", &id); 76 printf("%lld\n", f[id]); 77 } 78 } 79 } 80 81 int main() 82 { 83 #ifdef LOCAL 84 freopen("Test.in", "r", stdin); 85 #endif 86 87 Run(); 88 return 0; 89 }
L. Poor God Water
题意：
思路：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define mst(a,b) memset((a),(b),sizeof(a)) 4 5 typedef long long ll; 6 const int maxn = 10005; 7 const ll mod = 1e9 + 7; 8 const int INF = 0x3f3f3f3f; 9 const double eps = 1e-9; 10 11 12 ll fast_mod(ll a, ll n, ll Mod) 13 { 14 ll ans = 1; 15 a %= Mod; 16 while (n) 17 { 18 if (n & 1) ans = (ans*a) % Mod; 19 a = (a*a) % Mod; 20 n >>= 1; 21 } 22 return ans; 23 } 24 25 namespace Dup4 26 { 27 ll res[maxn], base[maxn], num[maxn], md[maxn]; 28 vector<int>vec; 29 void mul(ll *a, ll *b, int k) 30 { 31 for (int i = 0; i < 2 * k; i++) num[i] = 0; 32 for (int i = 0; i < k; i++) 33 { 34 if (a[i]) 35 { 36 for (int j = 0; j < k; j++) 37 { 38 num[i + j] = (num[i + j] + a[i] * b[j]) % mod; 39 } 40 } 41 } 42 for (int i = 2 * k - 1; i >= k; i--) 43 { 44 if (num[i]) 45 { 46 for (int j = 0; j < vec.size(); j++) 47 { 48 num[i - k + vec[j]] = (num[i - k + vec[j]] - num[i] * md[vec[j]]) % mod; 49 } 50 } 51 } 52 for (int i = 0; i < k; i++) a[i] = num[i]; 53 } 54 ll solve(ll n, vector<int> a, vector<int> b) 55 { 56 ll ans = 0, cnt = 0; 57 int k = a.size(); 58 assert(a.size() == b.size()); 59 for (int i = 0; i < k; i++) md[k - 1 - i] = -a[i]; 60 md[k] = 1; 61 vec.clear(); 62 for (int i = 0; i < k; i++) if (md[i]) vec.push_back(i); 63 for (int i = 0; i < k; i++) res[i] = base[i] = 0; 64 res[0] = 1; 65 while ((1LL << cnt) <= n) cnt++; 66 for (int p = cnt; p >= 0; p--) 67 { 68 mul(res, res, k); 69 if ((n >> p) & 1) 70 { 71 for (int i = k - 1; i >= 0; i--) res[i + 1] = res[i]; 72 res[0] = 0; 73 for (int j = 0; j < vec.size(); j++) 74 { 75 res[vec[j]] = (res[vec[j]] - res[k] * md[vec[j]]) % mod; 76 } 77 } 78 } 79 for (int i = 0; i < k; i++) ans = (ans + res[i] * b[i]) % mod; 80 if (ans < 0) ans += mod; 81 return ans; 82 } 83 vector<int> BM(vector<int> s) 84 { 85 vector<int> B(1, 1), C(1, 1); 86 int L = 0, m = 1, b = 1; 87 for (int i = 0; i < s.size(); i++) 88 { 89 ll d = 0; 90 for (int j = 0; j < L + 1; j++) d = (d + (ll)C[j] * s[i - j]) % mod; 91 if (d == 0) m++; 92 else if (2 * L <= i) 93 { 94 vector<int> T = C; 95 ll c = mod - d * fast_mod(b, mod - 2, mod) % mod; 96 while (C.size() < B.size() + m) C.push_back(0); 97 for (int j = 0; j < B.size(); j++) C[j + m] = (C[j + m] + c * B[j]) % mod; 98 L = i + 1 - L, B = T, b = d, m = 1; 99 } 100 else 101 { 102 ll c = mod - d * fast_mod(b, mod - 2, mod) % mod; 103 while (C.size() < B.size() + m) C.push_back(0); 104 for (int j = 0; j < B.size(); j++) C[j + m] = (C[j + m] + c * B[j]) % mod; 105 m++; 106 } 107 } 108 return C; 109 } 110 int gao(vector<int> a, ll n) 111 { 112 vector<int> c = BM(a); 113 c.erase(c.begin()); 114 for (int i = 0; i < c.size(); i++) c[i] = (mod - c[i]) % mod; 115 return solve(n, c, vector<int>(a.begin(), a.begin() + c.size())); 116 } 117 } 118 119 void RUN() 120 { 121 ll n; 122 int t; 123 scanf("%d", &t); 124 while(t--) 125 { 126 scanf("%lld", &n); 127 ++n; 128 printf("%d\n", Dup4::gao(vector<int>{3,9,20,46,106,244,560,1286,2956,6794}, n - 2)); 129 } 130 } 131 132 int main() 133 { 134 #ifdef LOCAL_JUDGE 135 freopen("Text.txt", "r", stdin); 136 #endif // LOCAL_JUDGE 137 138 RUN(); 139 140 #ifdef LOCAL_JUDGE 141 fclose(stdin); 142 #endif // LOCAL_JUDGE 143 144 145 return 0; 146 }
ACM-ICPC 2018 焦作赛区网络预赛 Solution
标签：res mes 多重 main for word put oom back
原文地址：https://www.cnblogs.com/Dup4/p/9651825.html
