A
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 3e5 + 5, MAXM = 6e6 + 5; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v, ll val) { to[++tot] = v; value[tot] = val; nxt[tot] = Head[u]; Head[u] = tot; }*/ int main() { ios_base::sync_with_stdio(0); cin.tie(0); int a, b, c, n; cin >> a >> b >> c >> n; if (max(c, max(a, b)) > n) { cout << -1 << endl; return 0; } if (c > min(a, b)) { cout << -1 << endl; return 0; } if (n - a - b + c < 1) { cout << -1 << endl; return 0; } cout << n - a - b + c << endl; return 0; }
B
*10后就不会有精度问题了
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 3e5 + 5, MAXM = 6e6 + 5; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v, ll val) { to[++tot] = v; value[tot] = val; nxt[tot] = Head[u]; Head[u] = tot; }*/ int num[105]; priority_queue<int, vector<int>, greater<int> > que; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n; cin >> n; int sum = 0; for (int i = 1; i <= n; i++) { cin >> num[i]; num[i] = num[i] * 10; que.push(num[i]); sum += num[i]; } int level = 45 * n; int ans = 0; while (sum < level) { sum += 50 - que.top(); que.pop(); ans++; } cout << ans << endl; return 0; }
C
一眼二分
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 3e5 + 5, MAXM = 6e6 + 5; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v, ll val) { to[++tot] = v; value[tot] = val; nxt[tot] = Head[u]; Head[u] = tot; }*/ int num[105]; priority_queue<int, vector<int>, greater<int> > que; ll n; ll check(ll x) { ll sum = 0; ll now = n; while (now > 0) { ll duce = min(x, now); sum += duce; now -= duce; now -= now / 10; } return sum; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> n; ll aim = n / 2 + n % 2; ll l = 0, r = 2 * n; ll mid; while (l < r - 1) { mid = (l + r) / 2; if (check(mid) >= aim) { r = mid; } else { l = mid; } } cout << r << endl; return 0; }
D
DP或者贪心模拟即可
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 3e5 + 5, MAXM = 6e6 + 5; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v, ll val) { to[++tot] = v; value[tot] = val; nxt[tot] = Head[u]; Head[u] = tot; }*/ char f[3][105]; char s[3][105]; string ch[3]; int n; int ans = 0; int main() { ios_base::sync_with_stdio(0); cin.tie(0); for (int i = 1; i <= 2; i++) { cin >> ch[i]; n = ch[i].size(); for (int j = 0; j < n; j++) { f[i][j + 1] = ch[i][j]; } } for (int i = 1; i <= 2; i++) { for (int j = 1; j <= n; j++) { s[i][j] = f[i][j]; } } int sumnow; for (int i = 1; i <= n - 1; i++) { sumnow = 0; for (int j = 1; j <= 2; j++) { for (int k = i; k <= i + 1; k++) { if (s[j][k] == '0') { sumnow++; } } } if (sumnow >= 3) { ans++; for (int j = 1; j <= 2; j++) { for (int k = i; k <= i + 1; k++) { if (s[j][k] == '0') { s[j][k] = 'X'; } } } if (sumnow == 4) { s[2][i + 1] = '0'; } } } cout << ans << endl; return 0; }
E
题意:
给你一个最多18位的数 你可以任意排列或者删除某个数 但是出现过的数至少要有一个且不能有前导零
问你有几种方案
解:
统计下每个数的个数 然后DFS一下算下答案即可
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 5e2 + 5, MAXM = 6e6 + 5; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v, ll val) { to[++tot] = v; value[tot] = val; nxt[tot] = Head[u]; Head[u] = tot; }*/ int num[MAXN]; int nownum[MAXN]; ll anser = 0; ll facnum[MAXN]; void init() { facnum[0] = 1; for (int i = 1; i <= 20; i++) { facnum[i] = i * facnum[i - 1]; } } ll calans () { ll ansnow = 0; for (int i = 0; i <= 9; i++) { ansnow += nownum[i]; } ansnow = facnum[ansnow]; for (int i = 0; i <= 9; i++) { ansnow /= facnum[nownum[i]]; } return ansnow; } void getans (int x) { if (x > 9) { anser += calans(); if (nownum[0] > 0) { nownum[0]--; anser -= calans(); nownum[0]++; } return ; } if (num[x]) { for (nownum[x] = 1; nownum[x] <= num[x]; nownum[x]++) { getans(x + 1); } } else { getans(x + 1); } } int main () { ios_base::sync_with_stdio(0); cin.tie(0); init(); ll pop; cin >> pop; while (pop > 0) { num[pop % 10]++; pop /= 10; } getans(0); cout << anser << endl; return 0; }