标签:geo 数学 https ring set nbsp end pid one
Problem A | 篮球队选拔 | 排序 |
Problem B | 不存在的泳池 | 水题 |
Problem C | 调酒壶里的酸奶 | 深搜 |
Problem D | 过分的谜题 | 模拟 |
Problem E | 小C的数学问题 | 单调栈 |
Problem F | fps游戏 | 数学题 |
Problem G | 闪闪发光 | 水题 |
Problem H | 流连人间的苏苏 | 水题 |
Problem I | 奔赴云南 | 水题 |
Problem J | TCMPC进阶之路 | 水题 |
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 200005; 5 int a[maxn], b[maxn], c[maxn]; 6 int main() { 7 int t; scanf("%d",&t); 8 while (t--) { 9 int n; scanf("%d",&n); 10 for (int i = 1; i <= 2*n; i++) scanf("%d",&a[i]); 11 for (int i = 1; i <= 2*n; i++) scanf("%d",&b[i]); 12 for (int i = 1; i <= 2*n; i++) c[i] = a[i]+b[i]; 13 sort(c+1,c+1+2*n); 14 if (c[n] < c[n+1]) printf("Cheat\n"); 15 else printf("Fail\n"); 16 } 17 return 0; 18 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() { 4 int a, b; 5 while (~scanf("%d%d",&a,&b)) { 6 int a2, a3, b2, b3, tmp; 7 a2 = a3 = b2 = b3 = 0; 8 tmp = a; 9 while (tmp % 2 == 0) tmp /= 2, a2++; 10 tmp = a; 11 while (tmp % 3 == 0) tmp /= 3, a3++; 12 tmp = b; 13 while (tmp % 2 == 0) tmp /= 2, b2++; 14 tmp = b; 15 while (tmp % 3 == 0) tmp /= 3, b3++; 16 if (a2 > b2) 17 for (int i = 1; i <= a2-b2; i++) a /= 2; 18 else if (b2 > a2) 19 for (int i = 1; i <= b2-a2; i++) b /= 2; 20 if (a3 > b3) 21 for (int i = 1; i <= a3-b3; i++) a /= 3; 22 else if (b3 > a3) 23 for (int i = 1; i <= b3-a3; i++) b /= 3; 24 if (a != b) puts("-1"); 25 else printf("%d\n",abs(a2-b2)+abs(a3-b3)); 26 } 27 return 0; 28 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 int a, b, c, ans; 4 const int inf = 0x3f3f3f3f; 5 bool vis[105][105]; 6 void dfs(int na, int nb, int cur) { 7 if (vis[na][nb]) return; 8 if (cur > ans) return; 9 if (na == c || nb == c) { 10 ans = min(ans,cur); 11 return; 12 } 13 vis[na][nb] = true; 14 dfs(a,nb,cur+1); 15 dfs(na,b,cur+1); 16 dfs(0,nb,cur+1); 17 dfs(na,0,cur+1); 18 dfs(max(0,na-(b-nb)),min(b,na+nb),cur+1); 19 dfs(min(a,nb+na),max(0,nb-(a-na)),cur+1); 20 vis[na][nb] = false; 21 } 22 int main() { 23 while (scanf("%d%d%d",&a,&b,&c) != EOF) { 24 if (c%__gcd(a,b) != 0) { 25 puts("impossible"); 26 continue; 27 } 28 memset(vis,0,sizeof(vis)); 29 ans = inf; 30 dfs(0,0,0); 31 printf("%d\n",ans); 32 } 33 return 0; 34 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 int a[10005]; 4 void get(int n) { 5 int pos = 1; pos += 1; 6 int ans = 1; 7 while (pos != 1) { 8 if (pos <= n) pos += pos, ans++; 9 else pos -= 2*n-pos+1, ans++; 10 } 11 a[n] = ans; 12 } 13 int main() { 14 for (int i = 1; i <= 10000; i++) 15 get(i); 16 int n; 17 while (scanf("%d",&n) != EOF) { 18 printf("%d\n",a[n]); 19 } 20 return 0; 21 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e5+5; 5 ll a[maxn], sum[maxn]; 6 int L[maxn], R[maxn]; 7 int main() { 8 int n; scanf("%d",&n); 9 for (int i = 1; i <= n; i++) { 10 scanf("%lld",&a[i]); 11 sum[i] = sum[i-1]+a[i]; 12 } 13 for (int i = 1; i <= n; i++) { 14 int l = i; 15 while (a[i] <= a[l-1] && l > 1) l = L[l-1]; 16 L[i] = l; 17 } 18 for (int i = n; i >= 1; i--) { 19 int r = i; 20 while (a[i] <= a[r+1] && r < n) r = R[r+1]; 21 R[i] = r; 22 } 23 ll ans = -1; 24 int l, r; 25 for (int i = 1; i <= n; i++) { 26 ll val = (sum[R[i]]-sum[L[i]-1])*a[i]; 27 if (ans < val) { 28 ans = val; 29 l = L[i]; 30 r = R[i]; 31 } 32 } 33 printf("%lld\n",ans); 34 printf("%d %d\n",l,r); 35 return 0; 36 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 const double pi = acos(-1); 4 int main() { 5 int d, r, c; double a; 6 scanf("%d%d%d%lf",&d,&r,&c,&a); 7 double v = atan(1.0*r/d)*180/pi; 8 int num = v/a + 1; 9 cout << max(0,c-num) << endl; 10 return 0; 11 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e6+1000; 5 ll a[maxn]; 6 int main() { 7 int n; 8 while (cin >> n) { 9 memset(a,0,sizeof(a)); 10 for (int i = 1; i <= n; i++) { 11 int x; scanf("%d",&x); 12 a[x]++; 13 } 14 ll ans = 0; 15 for (int i = 0; i <= 1000500; i++) { 16 a[i+1] += a[i]/2; 17 if (a[i] % 2 == 1) ans++; 18 } 19 printf("%lld\n",ans); 20 } 21 return 0; 22 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1010; 5 int n, d; 6 bool tree[maxn]; 7 void print() { 8 int l = 0; 9 vector<int> L, R; 10 for (int i = 1; i <= n+5; i++) { 11 if (tree[i] && l == 0) l = i; 12 if (!tree[i] && l != 0) { 13 L.push_back(l); 14 R.push_back(i-1); 15 l = 0; 16 } 17 } 18 printf("[%d,%d]",L[0],R[0]); 19 for (int i = 1; i < L.size(); i++) 20 printf(",[%d,%d]",L[i],R[i]); 21 printf("\n"); 22 } 23 int main() { 24 scanf("%d%d",&n,&d); 25 while (d--) { 26 int l, r; scanf("%d%d",&l,&r); 27 for (int i = l; i <= r; i++) tree[i] = true; 28 print(); 29 } 30 return 0; 31 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() { 4 string s; int t1, t2; 5 while (cin >> s >> t1 >> t2) { 6 t1 = t1/100*60 + t1%100; 7 t2 = t2/100*60 + t2%100; 8 printf("%s to Kunming: %02d:%02d\n",s.c_str(),(t2-t1)/60,(t2-t1)%60); 9 } 10 return 0; 11 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() { 4 int t; scanf("%d",&t); 5 while (t--) { 6 int n; scanf("%d",&n); 7 printf("%d\n",n*(n+1)/2); 8 } 9 return 0; 10 }
标签:geo 数学 https ring set nbsp end pid one
原文地址:https://www.cnblogs.com/wstong/p/11705932.html