B. Primal Sport
数学题，对 x2 和 x1 分解质因子即可。
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define rep(i,a,b) for (int i=a; i<=b; ++i) #define per(i,b,a) for (int i=b; i>=a; --i) #define mes(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define MP make_pair #define PB push_back #define fi first #define se second typedef long long ll; const int N = 200005; int p1[N], cnt1; void get(int x, int* p, int &num) { num = 0; for(int i=2; i<=sqrt(x); ++i) if(x%i==0) { p[++num] = i; while(x%i==0) x/=i; } if(x > 1) p[++num] = x; } int p2[N], cnt2; int x2; int main() { scanf("%d", &x2); get(x2, p1, cnt1); int ans = INF; rep(i,1,cnt1) { rep(x1, x2-p1[i]+1, x2) { get(x1, p2, cnt2); rep(l,1,cnt2) { if(x1-p2[l]+1 >= 3) ans = min(ans, x1-p2[l]+1); } } } printf("%d ", ans); return 0; }
C. Producing Snow
二分+树状数组，对于 v[i] 看它最后在哪天消耗完。 然后树状数组 区间更新，单点查询。
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define rep(i,a,b) for (int i=a; i<=b; ++i) #define per(i,b,a) for (int i=b; i>=a; --i) #define mes(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define MP make_pair #define PB push_back #define fi first #define se second typedef long long ll; const int N = 200005; int n; ll v[N], t[N], bit[N], sum[N], ans[N]; void Add(int x, ll y) { for( ; x<N; x+=x&-x) bit[x] += y; } ll Sum(int x) { ll ret=0; for( ; x; x-=x&-x) ret += bit[x]; return ret; } int main() { scanf("%d", &n); rep(i,1,n) scanf("%lld", &v[i]); rep(i,1,n) scanf("%lld", &t[i]), sum[i]=sum[i-1]+t[i]; sum[n+1] = 1e18; rep(i,1,n) { int pos = upper_bound(sum+1, sum+n+2, sum[i-1]+v[i]) - sum; ans[pos] += v[i] - (sum[pos-1]-sum[i-1]); Add(i, 1), Add(pos, -1); ans[i] += Sum(i)*t[i]; } rep(i,1,n) printf("%lld%c", ans[i], " "[i==n]); return 0; }
D. Perfect Security
字典树
拆分成二进制，先把第二个数组插入到 trie树里，再在 trie树里跑第一个数组中的数。有相同的位就取相同，没有就取相反的。
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define rep(i,a,b) for (int i=a; i<=b; ++i) #define per(i,b,a) for (int i=b; i>=a; --i) #define mes(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define MP make_pair #define PB push_back #define fi first #define se second typedef long long ll; const int N = 300005, M = 5e6; void get(int* num, int x) { rep(i,0,32) num[i] = 0; for(int len=1; x; x>>=1, ++len) { if(x&1) num[len] = 1; } } int n, A[N], Pi; int num[35], tr[M][2], val[M], tot; void Insert(int x) { get(num, x); for(int u=0, i=32; i>0; --i) { if(tr[u][num[i]] == 0) tr[u][num[i]] = ++tot; ++val[tr[u][num[i]]]; u = tr[u][num[i]]; } } int query(int x) { int ret = 0; get(num, x); for(int u=0, i=32; i>0; --i) { int tmp = num[i]^1; if(val[tr[u][num[i]]] > 0) { --val[tr[u][num[i]]]; u = tr[u][num[i]]; ret <<= 1, ret += num[i]; } else { --val[tr[u][tmp]]; u = tr[u][tmp]; ret <<= 1, ret += tmp; } } return ret; } int main() { scanf("%d", &n); rep(i,1,n) scanf("%d", &A[i]); rep(i,1,n) scanf("%d", &Pi), Insert(Pi); rep(i,1,n) printf("%d%c", query(A[i])^A[i], " "[i==n]); return 0; }