Building Blocks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

After 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 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, about 100 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 are n integers A1,A2,A3,……,An indicate
the height of each piles. (1≤Ai≤50000)

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 2
1 2 3 5
4 4 4
1 2 3 4

Sample Output

1
-1

Hint

In first case, LeLe move one block from third pile to first pile.

Source

BestCoder Round #34

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#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#define INF 1010010101000
#define ll long long

using 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;
        }
    }
}