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Lesha plays the recently published new version of the legendary game hacknet. In this version character skill mechanism was introduced. Now, each player character has exactly n skills. Each skill is represented by a non-negative integer ai — the current skill level. All skills have the same maximum level A.
Along with the skills, global ranking of all players was added. Players are ranked according to the so-called Force. The Force of a player is the sum of the following values:
Now Lesha has m hacknetian currency units, which he is willing to spend. Each currency unit can increase the current level of any skill by 1 (if it‘s not equal to A yet). Help him spend his money in order to achieve the maximum possible value of the Force.
The first line of the input contains five space-separated integers n, A, cf, cm and m (1 ≤ n ≤ 100 000, 1 ≤ A ≤ 109, 0 ≤ cf, cm ≤ 1000, 0 ≤ m ≤ 1015).
The second line contains exactly n integers ai (0 ≤ ai ≤ A), separated by spaces, — the current levels of skills.
On the first line print the maximum value of the Force that the character can achieve using no more than m currency units.
On the second line print n integers a‘i (ai ≤ a‘i ≤ A), skill levels which one must achieve in order to reach the specified value of the Force, while using no more than m currency units. Numbers should be separated by spaces.
3 5 10 1 5
1 3 1
12
2 5 2
3 5 10 1 339
1 3 1
35
5 5 5
In the first test the optimal strategy is to increase the second skill to its maximum, and increase the two others by 1.
In the second test one should increase all skills to maximum.
简单题意
你有n个技能可以学，你有m个技能点可以分配，每个技能的上限值都是A，每个技能都给出了至少要学的等级，然后你要合理分配技能点，使得战力最大
战力=满级技能*Cf+最小等级*Cm
然后我们枚举满级的技能个数，显然我们应该把那些至少学习等级大的点成满级，然后剩下的技能尽量最小等级最大
所以我们一开始从小到大排序，枚举后面i个变成满级，前面的二分最小等级，用前缀和判断可行性，然后计算出战力，更新答案
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 5 const int maxn=100100; 6 7 long long A,cf,cm,n,w,a[maxn],b[maxn],s[maxn],ans,tmp1,tmp2; 8 9 bool compare(long long x,long long y){ 10 return a[x]<a[y]; 11 } 12 13 long long find(long long x,long long rr){ 14 long long l,r,mid; 15 l=0;r=rr; 16 while(l!=r){ 17 mid=(l+r+1)/2; 18 if(a[b[mid]]<x)l=mid; 19 else r=mid-1; 20 } 21 return l; 22 } 23 24 int main(){ 25 scanf("%I64d%I64d%I64d%I64d%I64d",&n,&A,&cf,&cm,&w); 26 long long i; 27 for(i=1;i<=n;i++)scanf("%I64d",&a[i]); 28 for(i=1;i<=n;i++)b[i]=i; 29 sort(b+1,b+1+n,compare); 30 for(i=1;i<=n;i++)s[i]=s[i-1]+a[b[i]]; 31 w+=s[n]; 32 ans=-1; 33 for(i=1;i<=n+1;i++){ 34 if(A*(n-i+1)+s[i-1]>w)continue; 35 if(i==1)ans=cf*n+cm*A,tmp1=1; 36 if(i==1)break; 37 long long l=a[b[1]],r=A,mid,tt; 38 while(l!=r){ 39 mid=(l+r+1)/2; 40 tt=find(mid,i-1); 41 if(A*(n+1-i)+mid*tt+s[i-1]-s[tt]<=w)l=mid; 42 else r=mid-1; 43 } 44 if(ans<cf*(n+1-i)+l*cm)ans=cf*(n+1-i)+l*cm,tmp1=i,tmp2=l; 45 } 46 printf("%I64d\n",ans); 47 for(i=1;i<tmp1;i++) 48 if(a[b[i]]<tmp2)a[b[i]]=tmp2; 49 for(i=tmp1;i<=n;i++)a[b[i]]=A; 50 for(i=1;i<=n;i++)printf("%I64d ",a[i]); 51 return 0; 52 }
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原文地址：http://www.cnblogs.com/Randolph87/p/5199371.html
