题意：有N种棋子，M个人，已有K种收藏，要求最小的每个人送的棋子数使得最坏情况下至少有L种新的收藏，无解输出-1
N,M,K,L<=1e18
思路：
1 #include<cstdio> 2 #include<cstring> 3 #include<string> 4 #include<cmath> 5 #include<iostream> 6 #include<algorithm> 7 #include<map> 8 #include<set> 9 #include<queue> 10 #include<vector> 11 using namespace std; 12 typedef long long ll; 13 typedef unsigned int uint; 14 typedef unsigned long long ull; 15 typedef pair<int,int> PII; 16 typedef vector<int> VI; 17 #define fi first 18 #define se second 19 #define MP make_pair 20 #define N 1100000 21 #define MOD 1000000007 22 #define eps 1e-8 23 #define pi acos(-1) 24 25 26 int main() 27 { 28 ll n,m,k,l; 29 scanf("%lld%lld%lld%lld",&n,&m,&k,&l); 30 if(n<m||l>n-k) printf("-1 "); 31 else 32 { 33 ll ans=(k+l)/m; 34 if(ans*m<k+l) ans++; 35 if(ans*m<=n) printf("%lld ",ans); 36 else printf("-1 "); //7 4 1 6 ANS=-1 37 } 38 return 0; 39 }