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输出仅包含一行一个整数表示JYY可以宅的最多的天数。
【样例说明】
难得做回BZOJ
本题不确定买外卖的次数t,显然ans(t)为上凸函数，故用三分法
然后贪心，具体方法是，尽可能囤积最便宜(保质期够)的外卖，直至囤不了
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (200*2+10)
#define S first
#define P second
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
pair<ll,ll> a[MAXN],a2[MAXN];
ll m,f;
int n;
ll calc(ll c)
{
ll ans=0,M=m-c*f;
if (M<0) return 0;
int days=0;
For(i,n)
{
if (a[i].S+1<=days) continue;
ll days_block=a[i].S+1-days;
ll t=min(days_block,M/c/a[i].P);
ans+=c*t;
days+=t;
M-=c*t*a[i].P;
if (a[i].S+1<=days) continue;
ll single_day=min(c,M/a[i].P);
if (single_day>0)
{
ans+=single_day;
break;
}
}
return ans;
}
ll tri_search()
{
ll ans=0;
ll L=1,R=m/(f+a[1].P);
while(L<=R)
{
ll LM=L+(R-L)/3,RM=R-(R-L)/3;
ll Lans=calc(LM),Rans=calc(RM);
ans=max(max(Lans,Rans),ans);
if (Lans<Rans) L=LM+1;
else R=RM-1;
}
return ans;
}
int main()
{
// freopen("bzoj3874.in","r",stdin);
scanf("%lld%lld%d",&m,&f,&n);
For(i,n) scanf("%lld%lld",&a[i].P,&a[i].S);
sort(a+1,a+1+n);
memcpy(a2,a,sizeof(a));
int j=1;
Fork(i,2,n)
{
pair<ll,ll> &t=a2[i];
if (a[j].S==t.S&&a[j].P<t.P) continue;
while (j>0&&a[j].S<t.S&&a[j].P>=t.P) j--;
a[++j]=t;
}
n=j;
// For(i,n) cout<<a[i].S<<' '<<a[i].P<<endl;
printf("%lld\n",tri_search());
return 0;
}
BZOJ 3874([Ahoi2014]宅男计划-dp+三分法+贪心)
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原文地址：http://blog.csdn.net/nike0good/article/details/44752085
