题意:n个塔,第i个塔由$h_i$个cube组成,每次可以切去某高度h以上的最多k个cube,问你最少切多少次,可以让所有塔高度相等
k>=n, n<=2e5
思路:差分统计每个高度i有的方块数nh[i],然后从高到低贪心的切就行了
代码:
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> //#include<stack> #include<queue> #include<deque> #include<set> #include<vector> #include<map> #include<functional> #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PI; typedef pair<ll,ll> PLL; const db eps = 1e-6; const int mod = 100003; const int maxn = 2e5+100; const int maxm = 2e5+100; const ll inf = 0x3f3f3f3f3f3f3f3f; const db pi = acos(-1.0); int nh[maxn]; int main() { int n;ll m; scanf("%d %lld", &n, &m); int vol = 0; for(int i = 1; i <= n; i++){ int c; scanf("%d", &c); nh[0]++;nh[c+1]--; } for(int i = 1; i <= maxn; i++){ nh[i]+=nh[i-1]; } int tmp = 0; int ans = 0; for(int i = maxn; i >= 0; i--){ if(nh[i]==n)break; if(tmp+nh[i]>m){ tmp=nh[i]; ans++; } else tmp+=nh[i]; } if(tmp>0)ans++; printf("%d", ans); return 0; }
原文地址:https://www.cnblogs.com/wrjlinkkkkkk/p/9780373.html
时间: 2024-11-09 10:31:30