hihocoder #1499 : A Box of Coins 贪心

描述

Little Hi has a box which consists of 2xN cells as illustrated below.

+----+----+----+----+----+----+
| A1 | A2 | A3 | A4 | .. | AN |
+----+----+----+----+----+----+
| B1 | B2 | B3 | B4 | .. | BN |
+----+----+----+----+----+----+

There are some coins in each cell. For the first row the amounts of coins are A1, A2, ... AN and for the second row the amounts are B1, B2, ... BN.

Each second Little Hi can pick one coin from a cell and put it into an adjacent cell. (Two cells are adjacent if they share a common side. For example, A1 and A2 are adjacent; A1 and B1 are adjacent; A1 and B2 are not adjacent.)

Now Little Hi wants that each cell has equal number of coins by moving the coins. He wants to know the minimum number of seconds he needs to accomplish it.  

输入

The first line contains an integer, N. 2 <= N <= 100000  

Then follows N lines. Each line contains 2 integers Ai and Bi. (0 <= Ai, Bi <= 2000000000)  

It is guaranteed that the total amount of coins in the box is no more than 2000000000 and is a multiple of 2N.

输出

The minimum number of seconds.

样例输入

2
3 4
6 7

样例输出

4

.

题意 给出两行 A B 格子 里面放有一定数量的硬币

要求使得最后每个格子的硬币数量一样（输入保证合法）

移动每次只允许硬币在上下左右相邻的格子之间移动

求最少步数

贪心：

从最左边考虑，如果上下之间有一个有盈余，必定是相互补充最优。

其次 如果上下两个都有富余，则都往右边输送

再者 如果上下之间都缺乏，则都向左借

模拟即可  .

#include 
using namespace std;
typedef long long ll;

ll n,q;
ll A[2][100050];
int main()
{
     int n;
     cin>>n;
     ll tol=0;
     for(int i=1;i<=n;i++)
     {
         scanf("%lld%lld",&A[0][i],&A[1][i]);
        tol+=A[0][i]+A[1][i];
     }
     tol/=2*n;
     ll sum=0;
     ll nn=n;
     n=tol;
     for(int i=1;i<=nn;i++)
     {
         if (A[0][i]>n)        //superfluous
         {
             ll res=A[0][i]-n;
             if (A[1][i]0)
             {
                 ll tmp=min(res,(n-A[1][i]));
                 A[0][i]-=tmp;
                 sum+=tmp;
                 res-=tmp;
                 A[1][i]+=tmp;
             }
             if (res>0)
             {
                 A[0][i+1]+=res;
                 sum+=res;
                 A[0][i]=n;
             }
         }
         if (A[1][i]>n)
         {
             ll res=A[1][i]-n;
             if (A[0][i]0)
             {
                 ll tmp=min(res,(n-A[0][i]));
                 A[1][i]-=tmp;
                 sum+=tmp;
                 res-=tmp;
                 A[0][i]+=tmp;

             }
              if (res>0)
             {
                 A[1][i+1]+=res;
                 sum+=res;
                 A[1][i]=n;
             }
         }
         if (A[0][i]0)
             {
                 A[0][i+1]-=need;
                 sum+=need;
                 A[0][i]=n;
             }
         }
         if (A[1][i]0)
             {
                 A[1][i+1]-=need;
                 sum+=need;
                 A[1][i]=n;
             }
         }

     }
     printf("%lld\n",sum);

    return 0;
}