A：签到题，排序之后判断一下即可
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int N = 5e6 + 5; const LL Mod = 1e9 + 7; #define INF 1e9 #define dbg(x) cout << "now this num is " << x << endl; inline LL read() { LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();} return x * f; } int a[55],b[55]; int main() { int ca;ca = read(); while(ca--) { int n,x;n = read(),x = read(); for(int i = 1;i <= n;++i) a[i] = read(); for(int i = 1;i <= n;++i) b[i] = read(); sort(a + 1,a + n + 1); sort(b + 1,b + n + 1,greater<int>()); int f = 0; for(int i = 1;i <= n;++i) if(a[i] + b[i] > x) f = 1; printf("%s ",f ? "No" : "Yes"); } return 0; }
B：这题一开始没想通。
后面想了一下应该也就是各个边界的最值。
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int N = 5e6 + 5; const LL Mod = 1e9 + 7; #define INF 1e9 #define dbg(x) cout << "now this num is " << x << endl; inline LL read() { LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();} return x * f; } int main() { int ca;ca = read(); while(ca--) { int a,b,c,d;a = read(),b = read(),c = read(),d = read(); int ans = max(max(a + b,c + d),b + d); printf("%d ",ans); } return 0;
C：首先分类思考一下。
如果A < B,那么答案就是A，
如果A > B:
1.A % B != 0,答案还是A。
2.A % B == 0.
这种情况才是本题的难点，一开始觉得这个数就是1 ~ B里的最大因子，后面想了一下B~A里也有可能。
首先，如果X能整除B，那么显然X是B的倍数。
所以我们可以对B进行质因子分解，然后将某一个质因子少去一次，乘上A消去这个质因子所有次后的数
这样出来的值必定不可能是B的倍数，且尽可能大了，且是A的因子。
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<LL,int> pii; const int N = 1e6+5; const int M = 1e6+5; const LL Mod = 1e9+7; #define pi acos(-1) #define INF 1e18 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } } using namespace FASTIO; LL quick_mi(LL a,LL b) { LL re = 1; while(b) { if(b & 1) re = re * a; b >>= 1; a = a * a; } return re; } int main() { int ca;ca = read(); while(ca--) { LL p,q;p = read(),q = read(); LL ans = 1; if(p < q) ans = p; else if(p % q != 0) ans = p; else { int m = sqrt(q); for(int i = 2;i <= m;++i) { if(q % i == 0) { int cnt = 0; while(q % i == 0) q /= i,cnt++; LL ma = p; while(ma % i == 0) ma /= i; ans = max(ans,ma * quick_mi(i,cnt - 1)); } } if(q > 1) { LL ma = p; while(ma % q == 0) ma /= q; ans = max(ans,ma); } } printf("%lld ",ans); } system("pause"); return 0; }
D：先去数组进行一个排序，然后就可以证明：对于每个组合都是大的一半 的和 - 小的一半的和。
然后乘上组合方案数C（2n,n）即可
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<LL,int> pii; const int N = 2e5+5; const int M = 1e6+5; const LL Mod = 998244353; #define pi acos(-1) #define INF 1e18 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline LL read(){ LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } } using namespace FASTIO; LL a[N << 1],f[N << 1]; void init() { f[0] = 1; for(int i = 1;i < N * 2;++i) f[i] = f[i - 1] * i % Mod; } LL quick_mi(LL a,LL b) { LL re = 1; while(b) { if(b & 1) re = re * a % Mod; a = a * a % Mod; b >>= 1; } return re; } LL C(int n,int m) { return f[n] * quick_mi(f[m],Mod - 2) % Mod * quick_mi(f[n - m],Mod - 2) % Mod; } int main() { init(); int n;n = read(); for(int i = 1;i <= 2 * n;++i) a[i] = read(); sort(a + 1,a + 2 * n + 1); LL sum1 = 0,sum2 = 0; for(int i = 1;i <= n;++i) sum1 += a[i]; for(int i = n + 1;i <= 2 * n;++i) sum2 += a[i]; LL dis = (sum2 - sum1) % Mod; LL ans = dis * C(2 * n,n) % Mod; printf("%lld ",ans); system("pause"); return 0; }