题目链接：https://codeforces.com/contest/1183
A：
太傻不说。
B：
最大最小值之差跟2k比较一下就好了，也不说。
C：
算一下就好了，注意开long long。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 int q; 22 23 int main() { 24 cin >> q; 25 while (q--) { 26 ll k, n, a, b; cin >> k >> n >> a >> b; 27 if (n * b >= k) puts("-1"); 28 else printf("%lld ", min(n, (k - b * n - 1) / (a - b))); 29 } 30 return 0; 31 }
D：
倒着贪心就完事了。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 const int maxn = 2e5 + 10; 22 int q; 23 24 int main() { 25 scanf("%d", &q); 26 while (q--) { 27 int n, cnt[maxn] = {0}, maxx = 0; scanf("%d", &n); 28 rep1(i, 1, n) { 29 int x; scanf("%d", &x); maxx = max(maxx, x); 30 cnt[x]++; 31 } 32 sot(cnt, maxx); 33 ll ans = cnt[maxx]; int last = cnt[maxx]; 34 for (int i = maxx - 1; i; i--) { 35 if (!last || !cnt[i]) break; 36 if (cnt[i] >= last) ans += last - 1, last--; 37 else ans += cnt[i], last = cnt[i]; 38 } 39 printf("%lld ", ans); 40 } 41 return 0; 42 }
E：
bfs。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 int n, k; 22 ll ans = 0; 23 string s; 24 queue<string>q; 25 set<string>ss; 26 27 int main() { 28 cin >> n >> k >> s; 29 q.push(s); ss.insert(s); 30 while (!q.empty() && (int)ss.size() < k) { 31 auto curr = q.front(); 32 q.pop(); 33 rep1(i, 0, (int)curr.size()) { 34 string tmp = curr; 35 tmp.erase(i, 1); 36 if (ss.find(tmp) == ss.end() && ss.size() < k) { 37 q.push(tmp); ss.insert(tmp); ans += n - tmp.size(); 38 } 39 } 40 } 41 if ((int)ss.size() < k) puts("-1"); 42 else printf("%lld ", ans); 43 return 0; 44 }
F：
给定n个数，选出尽量多的数，在满足两两不能互相整除的情况下，和最大。
从大到小sort，然后O(n)扫一遍，选出最大的三个不能互质的数，再把和跟最大的数a/30*31取最大值即可。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 const int maxn = 2e5 + 10; 22 int q, a[maxn]; 23 24 int cmp(int a, int b) { 25 return a > b; 26 } 27 28 int main() { 29 scanf("%d", &q); 30 while (q--) { 31 int n; scanf("%d", &n); 32 rep1(i, 1, n) scanf("%d", &a[i]); 33 sort(a + 1, a + 1 + n, cmp); 34 ll ans = a[1], sec = 0, thir = 0; 35 int f1 = 0, f2 = 0, f3 = 0; 36 for (int i = 2; i <= n && !thir; i++) { 37 if (a[1] % a[i]) { 38 if (sec) { 39 if (sec % a[i]) { 40 thir = a[i]; 41 ans += a[i]; 42 } 43 } else { 44 sec = a[i]; 45 ans += a[i]; 46 } 47 } 48 if (a[i] * 2 == a[1]) f1 = 1; 49 else if (a[i] * 3 == a[1]) f2 = 1; 50 else if (a[i] * 5 == a[1]) f3 = 1; 51 } 52 if (f1 && f2 && f3) ans = max(ans, (ll)a[1] / 30 * 31); 53 printf("%lld ", ans); 54 } 55 return 0; 56 }
G && H：
摸了(