B. Checkout Assistant

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Bob came to a cash & carry store, put n items into his trolley, and went to the checkout counter to pay. Each item is described by its price ci and time ti in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant.

Input

The first input line contains number n (1 ≤ n ≤ 2000). In each of the following n lines each item is described by a pair of numbers ti, ci(0 ≤ ti ≤ 2000, 1 ≤ ci ≤ 109). If ti is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item i.

Output

Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay.

Examples

input

42 100 201 51 3

output

8

input

30 10 100 100

output

111

题意：买N件商品，每件商品有收钱时间ti，价格ci，收钱时可以偷窃其他ti件商品，求带走这N件商品需要支付的最小费用。

思路：可以理解为花ci的钱能得到ti+1件商品，于是用01背包做法dp[i]表示在这N件商品中带走i件支付的最小费用dp[i] = min( dp[i], dp[i-(ti+1)] + ci )。

http://blog.csdn.net/lvshubao1314/article/details/42505443

# include 
    
     # include 
     
      # include 
      
       using namespace std;int main(){    long long dp[2001], n, i, j, x, y;    while(~scanf("%I64d",&n))    {        memset(dp, 0x3f3f3f3f, sizeof(dp));        dp[0] = 0;        for(i=1; i<=n; ++i)        {            scanf("%I64d%I64d",&x, &y);            ++x;            for(j=n; j>=x; --j)                dp[j] = min(dp[j], dp[j-x]+y);            for(; j>0; --j) //这里表示当前物品数小于第i件物品的ti时，收费第i件物品时能偷走完j件。                dp[j] = min(dp[j], y);        }        printf("%I64d\n",dp[n]);    }    return 0;}