Codeforces 488C Fight the Monster【二分+枚举】

A. Fight the Monster

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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, ATKY - DEFM), while Yang's HP decreases by
max(0, ATKM - DEFY), 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 HPY, ATKY, DEFY, separated by a space,
denoting the initial HP,
ATK and DEF of Master Yang.

The second line contains three integers HPM, ATKM, DEFM, 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

99

Input

100 100 100
1 1 1
1 1 1

Output

0

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);
}
}