CodeForces 487A Fight the Monster

#include<iostream>
#include<cstdio>
using namespace std;
int judge(int hy,int ay,int dy,int hm,int am,int dm)//计算特定的攻击与防御之下,需要加多少hp
{
    if(am <= dy)
        return 0;
    int d1 = am - dy;
    //cout<<" d1 = "<<d1<<endl;
    int t1 = (hy + d1 - 1) / d1;
    //cout<<" t1 = "<<t1<<endl;
    int d2 = ay - dm;
    //cout<<" d2 = "<<d2<<endl;
    int t2 = (hm + d2 - 1) / d2;
    //cout<<" t2 = "<<t2<<endl;
    if(t1 > t2)
        return 0;
    int t = t2 * d1 - hy + 1;
    //cout<<" t = "<<t<<endl;
    return t;
}
int main()
{
    int hy,ay,dy,hm,am,dm,hp,ap,dp;
    cin>>hy>>ay>>dy;
    cin>>hm>>am>>dm;
    cin>>hp>>ap>>dp;
    long long ans = 0;
    if(dy >= am)  //y防御高过m攻击时,只要攻击高过防御即可
    {
        if(ay > dm)
            cout<<0<<endl;
        else
        {
            ans = ap * (dm - ay + 1);
            cout<<ans<<endl;
        }
    }
    else
    {
        long long ans0 = 0;//保证y攻击高过m防御需要的代价
        if(ay <= dm)
        {
            ans0 = ap * (dm - ay + 1);
            ay = dm + 1;
        }
        //cout<<"ans = "<<ans<<endl;
        ans = ans0 + judge(hy,ay,dy,hm,am,dm) * hp;
        //cout<<"ans = "<<ans<<endl;
        for(int i = 0; i*ap <= ans; i++) //枚举增加攻击防御的值并计算对应需要的hp,找最小值
        {
            for(int j = 0; i*ap + j*dp<= ans; j++)
            {
                long long tmp = ans0 + i * ap + j * dp + judge(hy,ay+i,dy+j,hm,am,dm) * hp;
                if(tmp < ans)
                {
                    //cout<<" i = "<<i<<" j = "<<j<<endl;
                    ans = tmp;
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

如果想要优化时间，那么考虑时必须要分析好血量、攻击力、防御力的变化
想做掉怪物首先需要保证本身的攻击力高过怪物的防御,然后枚举攻击力与防御力的增量,求出对应需要的hp,然后找出最小的值,数据范围都在100以内,因此枚举也可做,建议使用比较好分别的命名方式,否则用的时候各种串......

方法二：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define sc1(x) scanf("%s",&(x))
#define sc(x) scanf("%d",&(x))
#define sc3(x,y,z) scanf("%d%d%d", &(x), &(y), &(z))
#define pf(x) printf("%d
", x);
#define pf1(x) printf("%s
", x);
#define p(x) printf("%d ", x)
#define P printf("
")
#define pf2(x, y) printf("%d %d
", x, y)
#define FOR(i,b,e) for(int i=b; i<=e; i++)
#define FOR1(i,b,e) for(int i=b; i>=e; i--)
#define CL(x,y) memset(x,y,sizeof(x))
#define INF 0x7fffffff
int min(int x, int y)
{
        return x < y ? x : y;
}
int main()
{
    int wh, wa, wd, mh, ma, md, ch, ca, cd;
    sc3(wh, wa, wd);
    sc3(mh, ma, md);
    sc3(ch, ca, cd);
    int cost = INF;

    for(int da=0;da<=200;da++)
    {
        for(int dd=0; dd <=200;dd++)
        {
            for(int dh=0; dh<=1000;dh++)
            {
            if(wa+da-md <= 0) continue;
            int t = (mh-1) / (wa+da-md) + 1;
            if(dh < t*(ma-wd-dd)-wh+1)     continue;
            cost = min(cost, ca*da+cd*dd+ch*dh);
            }
        }
    }
    pf(cost);
    return 0;
}

直接暴力枚举，通过比较获取最小值

方法三：暴力+剪枝

#include <iostream>
using namespace std;
int main()
{
    int hy,ay,dy,hm,am,dm,h,a,d;
    cin>>hy>>ay>>dy>>hm>>am>>dm>>h>>a>>d;
    int i,j,k,ans=100000,sum,tmp;
    for(j=0;j<=max(0,dm+hm-ay);j++)
    {
        for(k=0;k<=max(0,am-dy);k++)
        {
            if(ay+j-dm<=0)
                break;
            if(ay+j-dm>0)
            {
                tmp=hm/(ay+j-dm);
                if(hm%(ay+j-dm)!=0)
                    tmp++;
                i=max(0,tmp*max(0,(am-dy-k))-hy+1);
            }
            sum=i*h+j*a+k*d;
           if(ans>sum)
                ans=sum;
        }
    }
    cout<<ans<<endl;
    return 0;
}