BC#34 1002 Hdu 5192 Building Blocks

Building BlocksTime limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): Accepted submission (s): 6


Problem Descriptionafter enjoying the Movie,lele went home alone. LeLe decided to build blocks.
LeLe has already built n Piles. He wants to move some blocks to make W Consecutive piles with exactly the same height H .

LeLe already put all of him 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 both piles already exists. For Instance,after one Move, "3 2 3" can become "2 2 4" or "3 2 2 1", and not "3 1 1 3".

You is request to calculate the minimum blocks should LeLe move.
Inputthere is multiple test cases, about - Cases.

The first line of input contains three integers N,W ,H (1≤N,W ,H ≤50000) . n Indicate n Piles blocks.

For the next line, there is n Integers A 1 , A 2 , A 3 ,... ..... .., A n Indicate the height of each piles. (1≤A i ≤ 50000 )

The height of a block is 1.
Outputoutput 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

1-1Hintin first case, LeLe move one block from third pile to first pile.

Sourcebestcoder Round #34
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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;            else Ans2+=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;                else Ans2-=h-a[first];                if (a[i]>h) ans1+=a[i]-h;                else Ans2+=h-a[i];                first++;                Ans=mmax (ANS1,ANS2);            Min=mmin (ans,min);        } cout<<min<<endl; }    }}

BC#34 1002 Hdu 5192 Building Blocks