Codeforces 488C Fight the Monster【二分+枚举】_a. fight the monster-优快云博客

主人公与怪物战斗，通过购买属性提升战斗力，目标是最小化花费确保胜利。文章介绍了一个算法，利用二分查找和枚举策略确定最低成本。

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A monster is attacking the Cyberland!

Master Yang, a braver, is going to beat the monster. Yang and the monster each have 3 attributes: hitpoints (HP), offensive power (ATK) and defensive power (DEF).

During the battle, every second the monster's HP decrease by max(0, ATK_Y - DEF_M), while Yang's HP decreases by max(0, ATK_M - DEF_Y), where index Y denotes Master Yang and index M denotes monster. Both decreases happen simultaneously Once monster's HP ≤ 0 and the same time Master Yang's HP > 0, Master Yang wins.

Master Yang can buy attributes from the magic shop of Cyberland: h bitcoins per HP, a bitcoins per ATK, and d bitcoins per DEF.

Now Master Yang wants to know the minimum number of bitcoins he can spend in order to win.

Input

The first line contains three integers HP_Y, ATK_Y, DEF_Y, separated by a space, denoting the initial HP, ATK and DEF of Master Yang.

The second line contains three integers HP_M, ATK_M, DEF_M, separated by a space, denoting the HP, ATK and DEF of the monster.

The third line contains three integers h, a, d, separated by a space, denoting the price of 1 HP, 1 ATK and 1 DEF.

All numbers in input are integer and lie between 1 and 100 inclusively.

Output

The only output line should contain an integer, denoting the minimum bitcoins Master Yang should spend in order to win.

Examples

Input

1 2 1
1 100 1
1 100 100

Output

Input

100 100 100
1 1 1
1 1 1

Output

Note

For the first sample, prices for ATK and DEF are extremely high. Master Yang can buy 99 HP, then he can beat the monster with 1 HP left.

For the second sample, Master Yang is strong enough to beat the monster, so he doesn't need to buy anything.

题目大意：

告诉你了主人公的血量，攻击力，防御值。

同时告诉你了敌人的血量，攻击力，防御值。

那么攻击一下敌人造成的伤害为攻击力-敌人的防御力

一个人的血量值低于

再告诉你了加一点血量的花费，加一点攻击力的花费，加一点防御值的花费。

问最少花费，使得主人公能够打败敌人。

思路：

1、明显有这样的递增性质：如果我们花费的钱越多，那么造成的伤害（胜利的机会）就越大。

那么我们二分花费。

2、对应二分出来的当前花费值mid，我们通过枚举可行解来判断：

①攻击力达到200+（因为敌人的防御力最高100，并且生命值最高200，我们极限枚举，一下就打死敌人的情况）

②防御力达到100+（因为敌人的攻击力最高100，这时候敌人就打不动我们了）

③生命值达到1000+（因为敌人的攻击力最高100，假设我们这个时候防御力为1，每秒能对敌人造成1点伤害，那么我们要抗住100*100回合才行）

全部考虑出极限值，然后进行枚举即可，判断当前二分值是否可行，同时记录可行解，一直二分下去即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
ll a,b,c;
ll d,e,f;
ll pa,pb,pc;
int Slove(ll money)
{
    for(ll i=0;i<=10250;i++)
    {
        if(pa*i>money)break;
        for(ll j=0;j<=220;j++)
        {
            if(pa*i+pb*j>money)break;
            for(ll k=0;k<=120;k++)
            {
                ll cost=pa*i+pb*j+pc*k;
                if(cost>money)break;
                ll tmpa=a+i;
                ll tmpb=b+j;
                ll tmpc=c+k;
                ll damage=tmpb-f;
                if(damage<=0)break;
                ll damage2=e-tmpc;
                if(damage2<=0)
                {
                    return 1;
                }
                ll win=d/damage;
                if(d%damage!=0)win++;
                ll win2=tmpa/damage2;
                if(tmpa%damage2!=0)win2++;
                if(win<win2)
                {
                    return 1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    while(~scanf("%I64d%I64d%I64d",&a,&b,&c))
    {
        scanf("%I64d%I64d%I64d",&d,&e,&f);
        scanf("%I64d%I64d%I64d",&pa,&pb,&pc);
        ll l=0;ll r=10000000000;
        ll ans=-1;
        while(r>=l)
        {
            ll mid=(l+r)/2;
            if(Slove(mid)==1)
            {
                r=mid-1;
                ans=mid;
            }
            else l=mid+1;
        }
        printf("%I64d\n",ans);
    }
}