Nastya and an Array
输出有几种不同的数字
#pragma comment(linker, "/STACK:102400000,102400000") #ifndef ONLINE_JUDGE #include "stdafx.h" #else #include<bits/stdc++.h> #endif using namespace std; typedef long long lint; typedef vector<int> VI; typedef pair<int, int> PII; typedef queue<int> QI; typedef map<int, int> MII; void makedata() { freopen("input.txt", "w", stdout); fclose(stdout); } int p[150000], q[150000]; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); int n, x; cin >> n; int ans = 0; for (int i = 0; i < n; i++) { cin >> x; if (x > 0) { if (!p[x]) ans++; p[x] = true; } if (x < 0) { if (!q[-x]) ans++; q[-x] = true; } } cout << ans << endl; return 0; }
Nastya Studies Informatics
把y/x分解质因数，a和b各自占有一部分且不重复
#pragma comment(linker, "/STACK:102400000,102400000") #ifndef ONLINE_JUDGE #include "stdafx.h" #else #include<bits/stdc++.h> #endif using namespace std; typedef long long lint; typedef vector<int> VI; typedef pair<int, int> PII; typedef queue<int> QI; typedef map<int, int> MII; void makedata() { freopen("input.txt", "w", stdout); fclose(stdout); } bool f[1000000]; int p[100]; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); memset(f, true, sizeof(f)); f[0] = f[1] = false; for (int i = 2; i < 1000000; i++) { for (int j = i + i; j < 1000000; j += i) f[j] = false; } int l, r, x, y; cin >> l >> r >> x >> y; if (y % x != 0) { cout << 0 << endl; return 0; } int m = y / x, sz = 0; for (int i = 2; i < 1000000; i++) { if (m == 1) break; if (!f[i]) continue; if (m % i == 0) { int tmp = 1; while (m % i == 0) { m /= i; tmp *= i; } p[sz++] = tmp; } } if (m != 1) p[sz++] = m; m = y / x; int ans = 0; for (int i = 0; i < (1 << sz); i++) { int a = 1, b; for (int j = 0; j < sz; j++) { if (i & (1 << j)) a *= p[j]; } b = m / a; if (l <= a * x && a * x <= r && l <= b * x && b * x <= r) ans++; } cout << ans << endl; return 0; }
Nastya and a Wardrobe
每次x2-0.5
#pragma comment(linker, "/STACK:102400000,102400000") #ifndef ONLINE_JUDGE #include "stdafx.h" #else #include<bits/stdc++.h> #endif using namespace std; typedef long long lint; typedef vector<int> VI; typedef pair<int, int> PII; typedef queue<int> QI; typedef map<int, int> MII; void makedata() { freopen("input.txt", "w", stdout); fclose(stdout); } inline lint pow(lint a, lint b, lint p) { lint rtn = 1; while (b) { if (b & 1) rtn = rtn * a % p; a = a * a % p; b >>= 1; } return rtn; } int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); lint x, k; cin >> x >> k; lint p = 1000000007; if (k == 0) { cout << (x * 2 % p) << endl; return 0; } if (x == 0) { cout << 0 << endl; return 0; } lint a = pow(2, k, p) * (x % p) % p, b = pow(2, k - 1, p); //a-b+0.5->2a-2b+1 cout << ((2 * a - 2 * b + 1) % p + p) % p << endl; return 0; } /* 2(2x - 0.5) - 0.5 2x - 0.5 1 4x - 1.5 2 8x - 3.5 4 16x - 7.5 8 32x - 15.5 16 */
Nastya and a Game
枚举左端点然后依次往右找，对于一段连续的1可以根据范围直接判断。因为S的最大值不超过2*10^18，所以向右找的次数不超过120次。
#pragma comment(linker, "/STACK:102400000,102400000") #ifndef ONLINE_JUDGE #include "stdafx.h" #else #include<bits/stdc++.h> #endif using namespace std; typedef long long lint; typedef vector<int> VI; typedef pair<int, int> PII; typedef queue<int> QI; typedef map<int, int> MII; void makedata() { freopen("input.txt", "w", stdout); fclose(stdout); } lint a[250000], s[250000]; int r[250000]; double ln[250000]; double inf = log(1LL << 60); int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); int n, k; cin >> n >> k; s[0] = 0; for (int i = 1; i <= n; i++) { cin >> a[i]; s[i] = s[i - 1] + a[i]; ln[i] = log(a[i]); } for (int i = n; i >= 1; i--) { if (a[i] == 1) { if (a[i + 1] == 1) r[i] = r[i + 1]; else r[i] = i; } } lint ans = 0; if (k == 1) ans = n; for (int i = 1; i <= n; i++) { int j = i + 1; lint m = a[i]; double p = ln[i]; while (j <= n) { if (a[j] == 1) { if (m % k == 0) { lint tmp = m / k; if (s[j] - s[i - 1] <= tmp && tmp <= s[r[j]] - s[i - 1]) ans++; } j = r[j] + 1; } else { if (p + ln[j] >= inf) break; p += ln[j]; m *= a[j]; if (m == (s[j] - s[i - 1])*k) ans++; j++; } } } cout << ans << endl; return 0; }
Nastya and King-Shamans
两个线段树，一个维护sum，一个维护max。对每个询问，如果a[1]=0直接输出1，否则令m=1，显然此时[1,m]里没有萨满王。对于m之后的，想成为萨满王其能力值至少为sum(1,m)，找到第一个满足条件的萨满p，判断p是不是萨满王，如果不是则令m=p，直到找到或确定没有。
#pragma comment(linker, "/STACK:102400000,102400000") #ifndef ONLINE_JUDGE #include "stdafx.h" #else #include<bits/stdc++.h> #endif using namespace std; typedef long long lint; typedef vector<int> VI; typedef pair<int, int> PII; typedef queue<int> QI; typedef map<int, int> MII; void makedata() { freopen("input.txt", "w", stdout); fclose(stdout); } lint a[250000]; int n, q; template <typename T> class SegmentTree { /* type: 0:sum 1:max */ private: T *data, *lazy; int type; void pushup(int rt) { if (type == 0) data[rt] = data[rt << 1] + data[rt << 1 | 1]; else if (type == 1) data[rt] = max(data[rt << 1], data[rt << 1 | 1]); } void pushdown(int rt, T m) { if (lazy[rt] == 0) return; lazy[rt << 1] += lazy[rt]; lazy[rt << 1 | 1] += lazy[rt]; if (type == 0) { data[rt << 1] += (m - (m >> 1)) * lazy[rt]; data[rt << 1 | 1] += (m >> 1) * lazy[rt]; } else if (type == 1) { data[rt << 1] += lazy[rt]; data[rt << 1 | 1] += lazy[rt]; } lazy[rt] = 0; } public: SegmentTree(int n, int t) : data((T *)malloc((n << 3) * sizeof(T))), lazy((T *)malloc((n << 3) * sizeof(T))), type(t) {} void Build(T * base, int l, int r, int rt) { lazy[rt] = 0; if (l == r) data[rt] = base[l]; else { int mid = (l + r) >> 1; Build(base, l, mid, rt << 1); Build(base, mid + 1, r, rt << 1 | 1); pushup(rt); } } void Modify(int l, int r, int rt, int L, int R, T v) { if (L <= l && R >= r) { lazy[rt] += v; if (type == 0) data[rt] += v * (r - l + 1); else if (type == 1) data[rt] += v; return; } pushdown(rt, r - l + 1); int mid = (l + r) >> 1; if (L <= mid) Modify(l, mid, rt << 1, L, R, v); if (R > mid) Modify(mid + 1, r, rt << 1 | 1, L, R, v); pushup(rt); } T QueryPoint(int l, int r, int rt, int val) { if (l == r) return data[rt]; pushdown(rt, r - l + 1); int mid = (l + r) >> 1; T ret = 0; if (val <= mid) ret = QueryPoint(l, mid, rt << 1, val); else ret = QueryPoint(mid + 1, r, rt << 1 | 1, val); pushup(rt); return ret; } T QuerySegment(int l, int r, int rt, int L, int R) { pushdown(rt, r - l + 1); if (L == l && R == r) return data[rt]; int mid = (l + r) >> 1; if (R <= mid) return QuerySegment(l, mid, rt << 1, L, R); if (mid < L) return QuerySegment(mid + 1, r, rt << 1 | 1, L, R); if (type == 0) return QuerySegment(l, mid, rt << 1, L, mid) + QuerySegment(mid + 1, r, rt << 1 | 1, mid + 1, R); else if (type == 1) return max(QuerySegment(l, mid, rt << 1, L, mid), QuerySegment(mid + 1, r, rt << 1 | 1, mid + 1, R)); } int find(int l, int r, int rt, int p, T x) { if (r <= p) return -1; if (data[rt] < x) return -1; if (l == r) return l; int mid = (l + r) >> 1; if (l > p) { if (data[rt << 1] >= x) return find(l, mid, rt << 1, p, x); else return find(mid + 1, r, rt << 1 | 1, p, x); } int rtn = find(l, mid, rt << 1, p, x); if (rtn != -1) return rtn; return find(mid + 1, r, rt << 1 | 1, p, x); } }; SegmentTree<lint> Tsum(250000, 0), Tmax(250000, 1); int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); cin >> n >> q; for (int i = 1; i <= n; i++) cin >> a[i]; Tsum.Build(a, 1, n, 1); Tmax.Build(a, 1, n, 1); while (q--) { int p; lint x; cin >> p >> x; Tsum.Modify(1, n, 1, p, p, x - a[p]); Tmax.Modify(1, n, 1, p, p, x - a[p]); a[p] = x; if (a[1] == 0) { cout << 1 << endl; continue; } int m = 1; while (1) { lint sum = Tsum.QuerySegment(1, n, 1, 1, m); int nex = Tmax.find(1,n,1,m,sum); if (nex == -1) { cout << -1 << endl; break; } if (a[nex] == Tsum.QuerySegment(1, n, 1, 1, nex - 1)) { cout << nex << endl; break; } m = nex; } } return 0; }