标签：water ros pillow int ide ted max ret src
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int INF = 0x3f3f3f3f; 4 const int maxn = 100 + 5; 5 typedef long long LL; 6 typedef pair<int, int>pii; 7 8 int main() 9 { 10 int n, m; 11 cin >> n >> m; 12 int d; 13 if(n == 1 || n == 3 || n == 5 || n == 7 || n == 8 || n == 10 || n == 12) 14 { 15 //31 16 d = 31 - (7 - m + 1); 17 } 18 else if(n == 2) 19 { 20 //28 21 d = 28 - (7 - m + 1); 22 } 23 else 24 { 25 d = 30 - (7 - m + 1); 26 } 27 cout << (d + 6) / 7 + 1 << endl; 28 29 return 0; 30 }
题意: 给你n个床,m个枕头.要求每个床最少分配一个枕头. 同时相邻的床的枕头个数之差要小于等于1; 要求第k张床的枕头数最大; 求这个最大值是多少.
思路：类似金字塔一样的一层一层叠上去。这题还可以用二分。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int INF = 0x3f3f3f3f; 4 const int maxn = 100 + 5; 5 typedef long long LL; 6 typedef pair<int, int>pii; 7 8 int main() 9 { 10 LL n, m, k; 11 while(cin >> n >> m >> k) 12 { 13 LL ans = 0; 14 m -= n, ans++; 15 LL leftlimit = k - 1, rightlimit = n - k; 16 LL left = 0, right = 0; 17 while(m > left + right) 18 { 19 ans ++; 20 m -= left + right + 1; 21 left = min(left + 1, leftlimit); 22 right = min(right + 1, rightlimit); 23 if(left == leftlimit && right == rightlimit) 24 { 25 ans += m / n; 26 break; 27 } 28 cout << ans << endl; 29 } 30 31 return 0; 32 }
Codeforces Round #393 (Div. 2)
标签：water ros pillow int ide ted max ret src
原文地址：http://www.cnblogs.com/luosuo10/p/6347867.html
