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Drying

Time Limit:?2000MS	?	Memory Limit:?65536K
Total Submissions:?9327	?	Accepted:?2364

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?≤ 10⁹). The third line contains?k?(1 ≤?k?≤ 10⁹).

Output

Output a single integer — the minimal possible number of minutes required to dry all clothes.

Sample Input

sample input #1 3 2 3 9 5sample input #2 3 2 3 6 5

Sample Output

sample output #1 3sample output #2 2

Source

Northeastern Europe 2005, Northern Subregion

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本來想貪心的，結果越寫越亂，老老實實二分。。。不過這題有其精妙之處，wa的人先看樣例

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送點test data給WA的朋友們……

Posted by MIRKING at 2008-07-31 13:39:06 on Problem 3104 and last updated at 2008-07-31 13:40:02

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test data： 2 8 9 2 result:6 7 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 2 result:875000000

Re:為什么二分時 枚舉的時間l,r,mid不用__int64就wa了

Posted by wangyaoxuan at 2010-10-14 11:30:23 on Problem 3104
In Reply To:為什么二分時 枚舉的時間l,r,mid不用__int64就wa了 Posted by:wangyaoxuan at 2010-10-14 11:22:39

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表示理解了,因為過程中求得的時間值t會超過int,后來的wa們慢慢體會....

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Posted by lgq1205 at 2009-12-03 12:54:33 on Problem 3104

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二分枚舉時間,判斷可行性,然后求解即可.注意判斷可行性的過程:1.先把所有衣服的含水量減去T 2.然后把>=(k-1)的拿去烘干，可以理解為烘干時候每分鐘掉(k-1)水，這樣所有的衣服都每分鐘自然干掉1水了。因為每分鐘掉一滴水是肯定的了，因此，如果你去烘干它的話，那么它就能再掉多k-1 + 1 == k，這樣才是k滴水,然后就是計算總花費時間 3.最后判斷總花費時間時候小于T,若小于等于,則可行性,否則不可行.

如果你WA了，看一下是否有變量要設為64位int，如果RE了，看一下除零。。。

Posted by 2745972075 at 2014-07-21 15:47:54 on Problem 3104

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1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<string> 12 13 #define N 100005 14 #define M 10 15 #define mod 1000000007 16 //#define p 10000007 17 #define mod2 1000000000 18 #define ll long long 19 #define LL long long 20 #define eps 1e-9 21 //#define inf 0x3fffffffff 22 #define maxi(a,b) (a)>(b)? (a) : (b) 23 #define mini(a,b) (a)<(b)? (a) : (b) 24 25 using namespace std; 26 27 int n; 28 ll ans; 29 ll a[N]; 30 ll k; 31 32 void ini() 33 { 34 ans=0; 35 int i; 36 for(i=1;i<=n;i++){ 37 scanf("%I64d",&a[i]); 38 } 39 scanf("%I64d",&k); 40 sort(a+1,a+1+n); 41 } 42 43 int ok(ll m) 44 { 45 int i; 46 ll le=m; 47 ll re; 48 for(i=1;i<=n;i++){ 49 re=a[i]-m; 50 if(re<=0) continue; 51 le-=(re+k-2)/(k-1); 52 } 53 if(le>=0) return 1; 54 return 0; 55 } 56 57 void solve() 58 { 59 ll l=1; 60 ll r=a[n]; 61 ll m; 62 //printf(" %d\n",a[1]); 63 if(k==1){ 64 ans=a[n];return; 65 } 66 while(l<r) 67 { 68 m=l+(r-l)/2; 69 if(ok(m)==1){ 70 r=m; 71 } 72 else{ 73 l=m+1; 74 } 75 } 76 ans=l; 77 } 78 79 void out() 80 { 81 printf("%I64d\n",ans); 82 } 83 84 int main() 85 { 86 // freopen("data.in","r",stdin); 87 //freopen("data.out","w",stdout); 88 //scanf("%d",&T); 89 //for(int ccnt=1;ccnt<=T;ccnt++) 90 //while(T--) 91 while(scanf("%d",&n)!=EOF) 92 { 93 ini(); 94 solve(); 95 out(); 96 } 97 98 return 0; 99 }

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轉載于:https://www.cnblogs.com/njczy2010/p/4211555.html

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