codeforces 488C Fight the Monster（枚举）

题目链接

C. 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 bymax(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.

Sample test(s)

input

1 2 11 100 11 100 100

output

99

input

100 100 1001 1 11 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.

题解：易知y的防御最大为100，y的攻击力最大为200，所以枚举y的防御和攻击力，计算出最少所需的血量，即可求出最小的花费。

代码如下：

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<string>#include<queue>#include<stack>#include<map>#include<set>#include<stdlib.h>#include<vector>#define inff 0x3fffffff#define nn 6100#define mod 1000000007typedef long long LL;const LL inf64=inff*(LL)inff;using namespace std;int hy,ay,dy;int hm,am,dm;int h,a,b;int main(){    int i,j;    while(scanf("%d%d%d",&hy,&ay,&dy)!=EOF)    {        scanf("%d%d%d",&hm,&am,&dm);        scanf("%d%d%d",&h,&a,&b);        int ans=inff;        int ix,fc;        for(i=dy;i<am;i++)        {            for(j=ay;j<=300;j++)            {                if(j<=dm)                    continue;                ix=(hm+j-dm-1)/(j-dm);                fc=ix*(am-i);                fc++;                fc=max(fc-hy,0);                ans=min(ans,fc*h+(j-ay)*a+(i-dy)*b);            }        }        ix=max(ay,dm+1);        fc=max(am-dy,0);        ans=min(ans,(ix-ay)*a+fc*b);        printf("%d\n",ans);    }    return 0;}