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傳送門Battle Ships

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Time Limit: 2 Seconds Memory Limit: 65536 KB

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Battle Ships is a new game which is similar to Star Craft. In this game, the enemy builds a defense tower, which has L longevity. The player has a military factory, which can produce N kinds of battle ships. The factory takes t_i seconds to produce the i

-th battle ship and this battle ship can make the tower loss l_i longevity every second when it has been produced. If the longevity of the tower lower than or equal to 0, the player wins. Notice that at each time, the factory can choose only one kind of battle ships to produce or do nothing. And producing more than one battle ships of the same kind is acceptable.

Your job is to find out the minimum time the player should spend to win the game.

Input

There are multiple test cases.
The first line of each case contains two integers N(1 ≤ N ≤ 30) and L(1 ≤ L ≤ 330), N is the number of the kinds of Battle Ships, L is the longevity of the Defense Tower. Then the following N

lines, each line contains two integers t _i(1 ≤ t _i ≤ 20) and l_i(1 ≤ l_i ≤ 330) indicating the produce time and the lethality of the i-th kind Battle Ships.

Output

Output one line for each test case. An integer indicating the minimum time the player should spend to win the game.

Sample Input

1 1
1 1
2 10
1 1
2 5
3 100
1 10
3 20
10 100

Sample Output

2
4
5

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Author: FU, Yujun
Contest: ZOJ Monthly, July 2012 【題目大意】有l滴血，你有n個戰艦，生產每個戰艦需要時間，生產後每秒攻擊i點血，求耗完l滴血的最少時間。 【code】

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int f[340],protim[40],fi[40];
int n,l,i,j;
int main()
{
    while(scanf("%d%d",&n,&l)!=EOF)
    {
        memset(f,0,sizeof(f));
        for(i=1;i<=n;i++)
        scanf("%d%d",&protim[i],&fi[i]); 
        for(i=1;i<=l;i++)
        for(j=1;j<=n;j++)
        f[i+protim[j]]=max(f[i+protim[j]],f[i]+i*fi[j]);
        for( i=1;i<=340;i++)
        if(f[i]>=l)break;
        printf("%d\n",i);
    }
    return 0;
}

Battle Ships（復習泛化物品**）