思路来自大佬jiangly

主要思路是dp

我理解并加上注释，供大家食用。

#include <cstdio>
#include <cstring>

#define int long long

using namespace std;

inline int read()
{
    int a=0,f=1;
    char ch=getchar();
    while(ch<48||ch>57)
    {
        if(ch==45)f=-1;
        ch=getchar();
    }
    while(48<=ch&&ch<=57)
    {
        a=(a<<1)+(a<<3)-48+ch;
        ch=getchar();
    }
    return a*f;
}

void write(int x)
{
    if(!x)return;
    write(x/10);
    putchar(x%10+48);
}

void pt(int x)
{
    if(!x)putchar(48);
    else if(x<0)x=-x,putchar(45);
    write(x);
    putchar(10);
}

const int N=5010;

int p1=read(),t1=read();
int p2=read(),t2=read();
int h=read(),s=read();
int c1=p1-s,c2=p2-s,c3=p1+p2-s;
int f[N];

int min(int a,int b)
{
    return a<b?a:b;
}

int max(int a,int b)
{
    return a>b?a:b; 
}

signed main()
{
    memset(f,0x3f,sizeof(f));
    f[0]=0;
    for(int i=1;i<=h;i++)
    {
        f[i]=min(f[max(0,i-c1)]+t1,f[max(0,i-c2)]+t2);  //即只选一个进行攻击 
        for(int j=1;j<=i;j++)                           //选择两个进行攻击 
        {
            if(j*t1>=t2)                                //最终时间取j*t1
            {
                int c=(j-1)*c1+(j*t1-t2)/t2*c2+c3;      //j-1:预留第一种lasers,j*t1-t2:预留第二种lasers 
                f[i]=min(f[i],f[max(0,i-c)]+j*t1); 
            }
            if(j*t2>=t1)
            {
                int c=(j-1)*c2+(j*t2-t1)/t1*c1+c3;
                f[i]=min(f[i],f[max(0,i-c)]+j*t2); 
            }
        }
    }
    pt(f[h]);
    return 0;
}