http://120.78.128.11/Problem.jsp?pid=3070
如果光是看技能伤害的话,就是一个简单的背包DP,但是现在有一个大招,怎么办呢,
对于这个大招进行分析,在L释放是temp这么多伤害,往后确是(x-L)*A+temp这么多伤害
对于一个时间段,那么如果每次只在L时间释放,也就是不蓄力,没有后续伤害,那么对于
时间段x,一共可以放x/L这么多次,剩下的时间也就是x%L拿来蓄力,这样会不会更好呢,
随着x的增大,这个值会随着x的增大,temp用的更多,最后的值更大,但是如果x超过一定
的值,肯定是蓄力伤害更高,所以考虑三分。
下面是三分的写法:
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;
#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e5+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}
inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}
void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;
LL p[maxx],dp[maxx];
int L,R,temp,A;
LL calc(int x)
{
int l=x/R,r=x/L;
W(r-l>3)
{
int mid=(l+r)/2;
int mmid=(mid+r)/2;
LL xx=min(1ll*mid*(R-L),1ll*(x-L*mid));
LL yy=min(1ll*mmid*(R-L),1ll*(x-L*mmid));
LL x1=1ll*mid*temp+xx*A;
LL x2=1ll*mmid*temp+yy*A;
if(x1>x2) r=mmid;
else l=mid;
}
LL ans=0;
FOR(l,r,i)
{
LL xx=min(1ll*i*(R-L),1ll*(x-L*i));
LL cc=1ll*i*temp+1ll*xx*A;
ans=max(cc,ans);
}
return ans;
}
void solve()
{
int t,x,a,y,b,z,c;
while(s_1(t)!=EOF)
{
me(dp,0);
me(p,0);
s_2(x,a);
s_2(y,b);
s_2(z,c);
s_2(L,R);
s_2(temp,A);
FOR(1,t,i)
p[i]=calc(i);
FOR(1,t,i)
{
if(i-x>=0)
dp[i]=max(dp[i-x]+a,dp[i]);
if(i-y>=0)
dp[i]=max(dp[i-y]+b,dp[i]);
if(i-z>=0)
dp[i]=max(dp[i-z]+c,dp[i]);
}
LL ans=0;
FOR(0,t,i)
ans=max(ans,dp[i]+p[t-i]);
print(ans);
}
}
int main()
{
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}