CodeForces 487A Fight the Monster

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 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
 5 {
 6     if(am <= dy)
 7         return 0;
 8     int d1 = am - dy;
 9     //cout<<" d1 = "<<d1<<endl;
10     int t1 = (hy + d1 - 1) / d1;
11     //cout<<" t1 = "<<t1<<endl;
12     int d2 = ay - dm;
13     //cout<<" d2 = "<<d2<<endl;
14     int t2 = (hm + d2 - 1) / d2;
15     //cout<<" t2 = "<<t2<<endl;
16     if(t1 > t2)
17         return 0;
18     int t = t2 * d1 - hy + 1;
19     //cout<<" t = "<<t<<endl;
20     return t;
21 }
22 int main()
23 {
24     int hy,ay,dy,hm,am,dm,hp,ap,dp;
25     cin>>hy>>ay>>dy;
26     cin>>hm>>am>>dm;
27     cin>>hp>>ap>>dp;
28     long long ans = 0;
29     if(dy >= am)  //y防御高过m攻击时,只要攻击高过防御即可
30     {
31         if(ay > dm)
32             cout<<0<<endl;
33         else
34         {
35             ans = ap * (dm - ay + 1);
36             cout<<ans<<endl;
37         }
38     }
39     else
40     {
41         long long ans0 = 0;//保证y攻击高过m防御需要的代价
42         if(ay <= dm)
43         {
44             ans0 = ap * (dm - ay + 1);
45             ay = dm + 1;
46         }
47         //cout<<"ans = "<<ans<<endl;
48         ans = ans0 + judge(hy,ay,dy,hm,am,dm) * hp;
49         //cout<<"ans = "<<ans<<endl;
50         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
51         {
52             for(int j = 0; i*ap + j*dp<= ans; j++)
53             {
54                 long long tmp = ans0 + i * ap + j * dp + judge(hy,ay+i,dy+j,hm,am,dm) * hp;
55                 if(tmp < ans)
56                 {
57                     //cout<<" i = "<<i<<" j = "<<j<<endl;
58                     ans = tmp;
59                 }
60             }
61         }
62         cout<<ans<<endl;
63     }
64     return 0;
65 }

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

方法二：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cmath>
 7 #include <stdio.h>
 8 #include <algorithm>
 9 #include <iostream>
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13 #define sc1(x) scanf("%s",&(x))
14 #define sc(x) scanf("%d",&(x))
15 #define sc3(x,y,z) scanf("%d%d%d", &(x), &(y), &(z))
16 #define pf(x) printf("%d\n", x);
17 #define pf1(x) printf("%s\n", x);
18 #define p(x) printf("%d ", x)
19 #define P printf("\n")
20 #define pf2(x, y) printf("%d %d\n", x, y)
21 #define FOR(i,b,e) for(int i=b; i<=e; i++)
22 #define FOR1(i,b,e) for(int i=b; i>=e; i--)
23 #define CL(x,y) memset(x,y,sizeof(x))
24 #define INF 0x7fffffff
25 int min(int x, int y)
26 {
27         return x < y ? x : y;
28 }
29 int main()
30 {
31     int wh, wa, wd, mh, ma, md, ch, ca, cd;
32     sc3(wh, wa, wd);
33     sc3(mh, ma, md);
34     sc3(ch, ca, cd);
35     int cost = INF;
36
37     for(int da=0;da<=200;da++)
38     {
39         for(int dd=0; dd <=200;dd++)
40         {
41             for(int dh=0; dh<=1000;dh++)
42             {
43             if(wa+da-md <= 0) continue;
44             int t = (mh-1) / (wa+da-md) + 1;
45             if(dh < t*(ma-wd-dd)-wh+1)     continue;
46             cost = min(cost, ca*da+cd*dd+ch*dh);
47             }
48         }
49     }
50     pf(cost);
51     return 0;
52 }

直接暴力枚举，通过比较获取最小值

方法三：暴力+剪枝

 1 #include <iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int hy,ay,dy,hm,am,dm,h,a,d;
 6     cin>>hy>>ay>>dy>>hm>>am>>dm>>h>>a>>d;
 7     int i,j,k,ans=100000,sum,tmp;
 8     for(j=0;j<=max(0,dm+hm-ay);j++)
 9     {
10         for(k=0;k<=max(0,am-dy);k++)
11         {
12             if(ay+j-dm<=0)
13                 break;
14             if(ay+j-dm>0)
15             {
16                 tmp=hm/(ay+j-dm);
17                 if(hm%(ay+j-dm)!=0)
18                     tmp++;
19                 i=max(0,tmp*max(0,(am-dy-k))-hy+1);
20             }
21             sum=i*h+j*a+k*d;
22            if(ans>sum)
23                 ans=sum;
24         }
25     }
26     cout<<ans<<endl;
27     return 0;
28 }