A.
全部空的放狗
B.
先O（NLOGNLOGN）处理出一个合数质因数中最大的质数是多少
因为p1 x1 x2的关系是 x2是p在x1之上的最小倍数 所以x1的范围是[x2-p+1,x2-1]要使最后答案尽可能小 要包含尽可能多的选择
p0 x0 x1关系同上
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; //const int maxn = 3e5 + 10; const int maxn = 100005; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; const int turn2[8][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}, {1, -1}, { -1, -1}, {1, 1}, { -1, 1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation ll mod = 3e7; int prime[1000005]; //priority_queue<int, vector<int>, less<int>> que; void init() { for (int i = 2; i <= 1000000; i++) { if (prime[i]) { continue; } for (int j = i * 2; j <= 1000000; j += i) { prime[j] = i; } } } int main() { int n; init(); cin >> n; int anser = INT_MAX; int now = prime[n]; //cout << prime[n] << endl; int x1 = n - now + 1; for (int i = x1; i <= n; i++) { if (!prime[i]) { continue; } anser = min(anser, i - prime[i] + 1); //cout << i - prime[i] + 1 << endl; } cout << anser << endl; }
C.
前缀和题
作温度的前缀和
二分出第i块雪在第j天融化
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation ll mod = 3e7; const int maxn=1e5+5; ll v[maxn]; ll t[maxn]; ll pre[maxn]; ll ans[maxn]; ll add[maxn]; int main() { ll n; cin >> n; for(int i=1;i<=n;i++) { scanf("%lld",v+i); } for(int i=1;i<=n;i++) { scanf("%lld",t+i); pre[i]=pre[i-1]+t[i]; } pre[n+1]=2e18+1; for(int i=1;i<=n;i++) { ll now=v[i]+pre[i-1]; int aim=upper_bound(pre,pre+n+1,now)-pre; //cout<<aim<<endl; if(aim==n+1) continue; add[aim]+=t[aim]-(pre[aim]-now); //cout<<t[aim]-(pre[aim]-now)<<endl; ans[aim]--; } for(int i=1;i<=n;i++) { ans[i]+=ans[i-1]; } for(int i=1;i<=n;i++) { ll anser=(ans[i]+i)*t[i]+add[i]; cout<<anser<<" "; } cout<<endl; }
D.01字典树带删除路径(用数组维护)
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation ll mod = 3e7; const int maxn = 3e5 + 5; int n, m; int ch[32 * maxn][2]; int sum[32 * maxn]; int a[maxn]; int b[maxn]; int node_cnt; inline void read(int &jqk) { jqk = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { jqk = (jqk << 3) + (jqk << 1) + c - '0'; c = getchar(); } jqk *= p; } void Insert(int x) { int cur = 0; for (int i = 30; i >= 0; i--) { int idx = (x >> i) & 1; if (!ch[cur][idx]) { //ch[node_cnt][1] = ch[node_cnt][0] = 0; ch[cur][idx] = ++node_cnt; } cur = ch[cur][idx]; sum[cur]++; } } int getans(int x) { int anser = 0; int cur = 0; for (int i = 30; i >= 0; i--) { int idx = (x >> i) & 1; if (!ch[cur][idx] || !sum[ch[cur][idx]]) { idx ^= 1; anser += (1 << i); } cur = ch[cur][idx]; sum[cur]--; } return anser; } int main() { int n; read(n); for (int i = 1; i <= n; i++) { read(a[i]); } for (int i = 1; i <= n; i++) { read(b[i]); Insert(b[i]); } for (int i = 1; i <= n; i++) { cout << getans(a[i]) << " "; } cout << endl; }