Codeforces Round #273 (Div. 2)
作者:互联网
A. Initial Bet
思路:
题目写的和shi一样,本来意思很简单,题目说的曲里拐弯的,就是看加起来能不能被5整除,由于都是正整数,所以要特判0的情况
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cmath> #include <queue> #define x first #define y second using namespace std; typedef long long LL; typedef pair<int, int>PII; const int N = 100010; const int MOD = 1000000007; int a[N]; int main() { int sum = 0; for (int i = 1; i <= 5; i++) cin >> a[i], sum += a[i]; int x = 0; x = sum / 5; if (sum % 5 != 0 || sum == 0) cout << -1 << endl; else cout << x << endl; return 0; }
B. Random Teams
思路:
当m-1个队人数为1,另一个队人数为n-m+1时分数最大,当各个队人数接近即 n%m个队人数为n/m+1,m-n%m个队人数为n/m时分数最小
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cmath> #include <queue> #define x first #define y second using namespace std; typedef long long LL; typedef pair<int, int>PII; const int N = 100010; const int MOD = 1000000007; int a[N]; int main() { LL n, m; scanf("%lld%lld", &n, &m); LL maxres = 0, minres = 0; maxres = (n - (m - 1)) * (n - (m - 1) - 1) / 2; //n%m个队是n/m+1,m-n%m个队是n/m minres = ((n % m) * ((n / m + 1) * (n / m) / 2)) + ((m - n % m) * ((n / m) * (n / m - 1) / 2)); printf("%lld %lld\n", minres, maxres); return 0; }
C. Table Decorations
思路:
不妨设a<b<c,则当c>=2*(a+b)时答案ans=a+b,当c<2*(a+b)时答案res=(a+b+c)/3
这题记住结论得了,感觉结论还挺有用的
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cmath> #include <queue> #define x first #define y second using namespace std; typedef long long LL; typedef pair<int, int>PII; const int N = 100010; const int MOD = 1000000007; LL a[4]; int main() { for (int i = 1; i <= 3; i++) cin >> a[i]; sort(a + 1, a + 1 + 3); LL res = 0; if (a[3] >= 2 * (a[1] + a[2])) res = a[1] + a[2]; else res = (a[1] + a[2] + a[3]) / 3; cout << res << endl; return 0; }
标签:typedef,const,int,LL,Codeforces,long,273,Div,include 来源: https://www.cnblogs.com/yctql/p/15388739.html