Building BlocksTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 60Accepted Submission(s): 6
Problem Description
After enjoying the movie,LeLe went home alone. LeLe decided to build blocks.LeLe has already builtpiles. He wants to move some blocks to makeconsecutive piles with exactly the same height.LeLe already put all of his blocks in these piles, which means he can not add any blocks into them. Besides, he can move a block from one pile to another or a new one，but not the position betweens two piles already exists.For instance,after one move,"3 2 3" can become "2 2 4" or "3 2 2 1",but not "3 1 1 3".You are request to calculate the minimum blocks should LeLe move.
Input
There are multiple test cases, aboutcases.The first line of input contains three integerspiles blocks.For the next line ,there areindicate the height of each piles.The height of a block is 1.
Output
Output the minimum number of blocks should LeLe move.If there is no solution, output "-1" (without quotes).
Sample Input
4 3 21 2 3 54 4 41 2 3 4
Sample Output
1-1
Hint
In first case, LeLe move one block from third pile to first pile.
Source
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#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cstdlib>#define INF 1010010101000#define ll long longusing namespace std;int n,m,h;ll a[50020*3];ll mmin(ll a,ll b) {if(a>b)return b;return a;}ll mmax(ll a,ll b) {if(a<b)return b;return a;}int main() {while(cin>>n>>m>>h) {ll sum=0;for(int i=1; i<=m; i++)a[i]=0;for(int i=m+1; i<=m+n; i++) {scanf("%I64d",&a[i]);sum+=a[i];}for(int i=m+n+1; i<=n+m+m; i++)a[i]=0;ll ans=0;if(sum<(ll)h*m)printf("-1\n");else {int first=1;ll Min=INF;ll ans1=0;ll ans2=0;for(int i=1; i<=m; i++) {if(a[i]>h)ans1+=a[i]-h;elseans2+=h-a[i];}ans=max(ans1,ans2);Min=ans;for(int i=m+1; i<=m+m+n; i++) {if(a[first]>h)ans1-=a[first]-h;elseans2-=h-a[first];if(a[i]>h)ans1+=a[i]-h;elseans2+=h-a[i];first++;ans=mmax(ans1,ans2);Min=mmin(ans,Min);}cout<<Min<<endl;}}}
，为你的难过而快乐的是敌人，
