标签：play pre over put scanf ace amp %s img
cf818A
1 #include <bits/stdc++.h> 2 #define ll __int64 3 using namespace std; 4 5 ll n, k; 6 7 int main() { 8 cin >> n >> k; 9 ll a = n / 2 / (k+1); 10 ll b = k * a; 11 printf("%I64d %I64d %I64d\n", a, b, n-a-b); 12 return 0; 13 }
818B，没有指示的数字随便赋值即可。
1 #include <bits/stdc++.h> 2 #define ll __int64 3 using namespace std; 4 5 int n, m; 6 int l[105]; 7 int a[105]; 8 bool mark[105]; 9 vector<int> v; 10 11 int main() { 12 cin >> n >> m; 13 for (int i = 0; i < m; i++) { 14 cin >> l[i]; 15 int k = (l[i] - l[i-1] + n) % n; 16 if (!k) k = n; 17 if (!a[l[i-1]]) { 18 a[l[i-1]] = k; 19 } else if (a[l[i-1]] != k) { 20 puts("-1"); 21 return 0; 22 } 23 } 24 25 for (int i = 1; i <= n; i++) { 26 if (a[i]) { 27 if (!mark[a[i]]) 28 mark[a[i]] = true; 29 else { 30 puts("-1"); 31 return 0; 32 } 33 } else { 34 v.push_back(i); 35 } 36 } 37 for (int i = 1; i <= n; i++) { 38 if (!mark[i]) { 39 a[v.back()] = i; 40 v.pop_back(); 41 } 42 } 43 44 for_each(a+1, a+1+n, [](int x) {cout << x << " ";}); 45 return 0; 46 }
818C，sweep line，坐标点为下标差分计数。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxcor = 1e5 + 5; 5 int d, n, m; 6 int cntl, cntr, cntt, cntb; 7 struct Point { 8 int x1, y1, x2, y2; 9 }sofa[maxcor]; 10 int cnt_left[maxcor], cnt_right[maxcor], cnt_top[maxcor], cnt_bottom[maxcor]; 11 12 inline bool l(int i) { 13 int k = cnt_left[max(sofa[i].x1, sofa[i].x2) - 1]; 14 if (sofa[i].x1 != sofa[i].x2) k--; 15 return k == cntl; 16 } 17 18 inline bool r(int i) { 19 int k = cnt_right[min(sofa[i].x1, sofa[i].x2) + 1]; 20 if (sofa[i].x1 != sofa[i].x2) k--; 21 return k == cntr; 22 } 23 24 inline bool t(int i) { 25 int k = cnt_top[max(sofa[i].y1, sofa[i].y2) - 1]; 26 if (sofa[i].y1 != sofa[i].y2) k--; 27 return k == cntt; 28 } 29 30 inline bool b(int i) { 31 int k = cnt_bottom[min(sofa[i].y1, sofa[i].y2) + 1]; 32 if (sofa[i].y1 != sofa[i].y2) k--; 33 return k == cntb; 34 } 35 36 int main() { 37 cin >> d >> n >> m; 38 for (int i = 1; i <= d; i++) { 39 int x1, y1, x2, y2; 40 cin >> x1 >> y1 >> x2 >> y2; 41 sofa[i] = {x1, y1, x2, y2}; 42 43 cnt_left[min(x1, x2)]++; 44 cnt_right[max(x1, x2)]++; 45 cnt_top[min(y1, y2)]++; 46 cnt_bottom[max(y1, y2)]++; 47 } 48 cin >> cntl >> cntr >> cntt >> cntb; 49 50 for (int i = 1; i <= 1e5; i++) { 51 cnt_left[i] += cnt_left[i-1]; 52 cnt_top[i] += cnt_top[i-1]; 53 } 54 for (int i = 1e5; i; i--) { 55 cnt_right[i] += cnt_right[i+1]; 56 cnt_bottom[i] += cnt_bottom[i+1]; 57 } 58 59 for (int i = 1; i <= d; i++) { 60 if (l(i) && r(i) && t(i) && b(i)) { 61 cout << i << endl; 62 return 0; 63 } 64 } 65 puts("-1"); 66 return 0; 67 }
818D，gameover之时将此数打标记，否则继续计数。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxcolor = 1e6 + 5; 5 bool mark[maxcolor]; 6 int n, A, cntA; 7 int cnt[maxcolor]; 8 9 int main() { 10 cin >> n >> A; 11 mark[A] = true; 12 for (int i = 1; i <= n; i++) { 13 int a; 14 scanf("%d", &a); 15 if (a == A) cntA++; 16 else { 17 if (mark[a]) 18 continue; 19 else if (cnt[a] < cntA) 20 mark[a] = true; 21 else cnt[a]++; 22 } 23 } 24 for (int i = 1; i <= 1e6; i++) 25 if (!mark[i] && cnt[i] >= cntA) { 26 cout << i << endl; 27 return 0; 28 } 29 puts("-1"); 30 return 0; 31 }
BZOJ1011，鬼题勿做。思路根本就不是在求正解，以及迷之SPJ。
1 #include <cstdio> 2 #define db double 3 4 const int maxn = 1e5 + 5; 5 int n; 6 db A; 7 db m[maxn], sum[maxn]; 8 9 int main() { 10 scanf("%d%lf", &n, &A); 11 for (int i = 1; i <= n; i++) { 12 scanf("%lf", &m[i]); 13 sum[i] = sum[i-1] + m[i]; 14 } 15 for (int i = 1; i <= n; i++) { 16 int k = int(A * db(i) + 1e-8); 17 db ans = 0.0; 18 if (k <= 100) { 19 for (int j = 1; j <= k; j++) 20 ans += m[i] * m[j] / (i-j); 21 } else { 22 ans = sum[k] * m[i] / (i-k/2); 23 } 24 printf("%.6lf\n", ans); 25 } 26 return 0; 27 }
POJ3468，分块。
1 #include <cstdio> 2 #include <cmath> 3 #define ll long long 4 #define maxn 100005 5 #define rep(i, x, y) for (int i = x; i <= y; i++) 6 7 int n, m, t; 8 int L[maxn], R[maxn], pos[maxn]; 9 ll a[maxn]; 10 ll sum[maxn], add[maxn]; 11 12 void Add(int l, int r, int k) { 13 int p = pos[l], q = pos[r]; 14 if (p == q) { 15 rep(i, l, r) a[i] += k; 16 sum[p] += k * (r - l + 1); 17 } else { 18 rep(i, p+1, q-1) add[i] += k; 19 rep(i, l, R[p]) a[i] += k; 20 sum[p] += k * (R[p] - l + 1); 21 rep(i, L[q], r) a[i] += k; 22 sum[q] += k * (r - L[q] + 1); 23 } 24 } 25 26 ll query(int l, int r) { 27 int p = pos[l], q = pos[r]; 28 ll ans = 0ll; 29 if (p == q) { 30 rep(i, l, r) ans += a[i]; 31 ans += add[p] * (r - l + 1); 32 } else { 33 rep(i, p+1, q-1) ans += sum[i] + add[i] * (R[i] - L[i] + 1); 34 rep(i, l, R[p]) ans += a[i]; 35 ans += add[p] * (R[p] - l + 1); 36 rep(i, L[q], r) ans += a[i]; 37 ans += add[q] * (r - L[q] + 1); 38 } 39 return ans; 40 } 41 42 int main() { 43 scanf("%d%d", &n, &m); 44 rep(i, 1, n) scanf("%lld", &a[i]); 45 46 t = sqrt(n); 47 rep(i, 1, t) { 48 L[i] = (i-1) * t + 1; 49 R[i] = i * t; 50 } 51 if (R[t] < n) { 52 t++; 53 L[t] = R[t-1] + 1; 54 R[t] = n; 55 } 56 rep(i, 1, t) { 57 rep(j, L[i], R[i]) { 58 pos[j] = i; 59 sum[i] += a[j]; 60 } 61 } 62 63 rep(i, 1, m) { 64 char ch[3]; 65 int a, b; 66 scanf("%s%d%d", ch, &a, &b); 67 if (ch[0] == ‘C‘) { 68 int c; 69 scanf("%d", &c); 70 Add(a, b, c); 71 } else { 72 printf("%lld\n", query(a, b)); 73 } 74 } 75 76 return 0; 77 }
标签：play pre over put scanf ace amp %s img
原文地址：https://www.cnblogs.com/AlphaWA/p/10363662.html
