Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 15512 | Accepted: 3942 |
Description
It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring process. Jane has decided to use a radiator to make drying faster. But the radiator is small, so it can hold only one thing at a time.
Jane wants to perform drying in the minimal possible time. She asked you to write a program that will calculate the minimal time for a given set of clothes.
There are n clothes Jane has just washed. Each of them took a_{i} water during washing. Every minute the amount of water contained in each thing decreases by one (of course, only if the thing is not completely dry yet). When amount of water contained becomes zero the cloth becomes dry and is ready to be packed.
Every minute Jane can select one thing to dry on the radiator. The radiator is very hot, so the amount of water in this thing decreases by k this minute (but not less than zero — if the thing contains less than k water, the resulting amount of water will be zero).
The task is to minimize the total time of drying by means of using the radiator effectively. The drying process ends when all the clothes are dry.
Input
The first line contains a single integer n (1 ≤ n ≤ 100 000). The second line contains a_{i} separated by spaces (1 ≤ a_{i} ≤ 10^{9}). The third line contains k (1 ≤ k ≤ 10^{9}).
Output
Output a single integer — the minimal possible number of minutes required to dry all clothes.
Sample Input
1 2 3 4 5 6 7 8 9 |
<b>sample input #1</b> 3 2 3 9 5 <b>sample input #2</b> 3 2 3 6 5 |
Sample Output
1 2 3 4 5 |
<b>sample output #1</b> 3 <b>sample output #2</b> 2 |
分析:
这道题,非常傻逼.
这道题就是二分时间,二分这个自由蒸发的时间。然后再进行判断就可以水过了/
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 |
#include<cstdio> #include<algorithm> #include<math.h> using namespace std; long long int w[100010]; long long int n,k; long long int max_mean; long long int ans; int main() { while(~scanf("%lld",&n)) { ans=0;max_mean=0;k=0; for(long long int i=1;i<=n;++i) { scanf("%lld",&w[i]); max_mean=max(w[i],max_mean); } scanf("%lld",&k); long long int left=0,right=max_mean+1,mid; if(k==1) { printf("%lld\n",max_mean); continue; } while(left<=right) { mid=(left+right)>>1; long long int sum=0; for(long long int i=1;i<=n;++i) { if(w[i]<mid)continue; sum+=ceil((double)(w[i]-mid)/(k-1)); } if(sum<=mid) { ans=mid; right=mid-1; } else left=mid+1; } printf("%lld\n",ans); } return 0; } |