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BestCoder Round #34B——数学——Building Blocks

分类: IT文章 • 2022-03-30 18:09:19

BestCoder Round #34B——数学——Building Blocks

Problem Description

After enjoying the movie,LeLe went home alone. LeLe decided to build blocks. LeLe has already built 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 piles blocks.

For the next line ,there are )

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.
大意：搭积木，连续的要3个，高度要是2，共4组，每次放能放到左边或者右边，不能放到两组之间，开long long 
把区间移动
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int main()
{
    int n,w,h;
    int a[110];
    while(~scanf("%d%d%d",&n,&w,&h)){
        ll l ,r, sum;
        l = r = sum =0;
        memset(a,0,sizeof(a));
            for(int i = w + 1; i <= w + n;i++){
                scanf("%d",&a[i]);
                sum += a[i];
            }
            if(sum < w*h){
                    printf("-1
");
                    continue;
            }
            int min1 = inf;
            for(int i = 1; i <= 2*w + n;i++){
             if(a[i] - h > 0)
             r += a[i] - h;
             else  l += a[i] - h;
             if(i >= w){
                  l = -l;
                 int temp = max(l,r);
                 min1 = min(min1,temp);
             l = -l;
            if(a[i-w+1]-h > 0)
            r -= a[i-w+1]-h;
            else l -= a[i-w+1] - h;
           }
        }
        printf("%d
",min1);
    }
    return 0;
}
View Code

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