Codeforces 371C Hamburgers（二分法）

骑猪看日落 2022-01-23 11:01 88阅读 0赞

Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite “Le Hamburger de Polycarpus” as a string of letters ‘B’ (bread), ‘S’ (sausage) и ‘C’ (cheese). The ingredients in the recipe go from bottom to top, for example, recipe “ВSCBS” represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.  
Polycarpus has nb pieces of bread, ns pieces of sausage and nc pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are pb rubles for a piece of bread, ps for a piece of sausage and pc for a piece of cheese.  
Polycarpus has r rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.

**Input**  
The first line of the input contains a non-empty string that describes the recipe of “Le Hamburger de Polycarpus”. The length of the string doesn’t exceed 100, the string contains only letters ‘B’ (uppercase English B), ‘S’ (uppercase English S) and ‘C’ (uppercase English C).  
The second line contains three integers nb, ns, nc (1 ≤ nb, ns, nc ≤ 100) — the number of the pieces of bread, sausage and cheese on Polycarpus’ kitchen. The third line contains three integers pb, ps, pc (1 ≤ pb, ps, pc ≤ 100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer r (1 ≤ r ≤ 1012) — the number of rubles Polycarpus has.  
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

**Output**  
Print the maximum number of hamburgers Polycarpus can make. If he can’t make any hamburger, print 0.

**Examples**

**Input**  
BBBSSC  
6 4 1  
1 2 3  
4

**Output**

**Input**  
BBC  
1 10 1  
1 10 1  
21

**Output**  
7

**Input**  
BSC  
1 1 1  
1 1 3  
1000000000000

**Output**  
200000000001

题意：每次制作一个汉堡需要给定的制作数量由字符串决定，用三个变量统计下，  
写一个公式然后用二分查找就ac了。

公式推理：  
每次制作b需要n1个，s需要n2个，c需要n3个，假设制作num个（num有二分出来的值），

制作b需要 max（0，num \* n1-nb）\* pb 钱  
制作s需要 max（0，num \* n2-ns）\* ps 钱  
制作c需要 max（0，num \* n3-nc）\* pc 钱

num \* n1 是制作num个汉堡需要b材料一共的数量 减去nb是减去原来有的材料， 乘上单价pb 注意，可能原来的材料已经够制作汉堡得了，所以不需要花钱公式计算出来肯定是个负数，所以要取个最大值0  
材料s c 同理…

二分查找： （取右区间要取得比最大值要大）

while(l<=r)
    {
    	mid=(l+r)>>1;
    	if(getsum(mid)==n)
    	{
    		cout<<mid<<endl;
    		flag=1; 
    		break;
    	}
    	if(getsum(mid)<n)
    	{
    		l=mid+1;
    		ans=mid;
    	}
    	else
    	{
    		r=mid-1;
    	}
    }

AC代码：

#include<iostream>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    const ll mmax=1e13;
    ll n1,n2,n3,nb,ns,nc,pb,ps,pc,sum,n,mid;
    ll getsum(ll num)
    {
    	return (ll)max((ll)0,num*n1-nb)*pb+max((ll)0,num*n2-ns)*ps+max((ll)0,num*n3-nc)*pc;
    }
    int main()
    {
    	string s;
    	while(cin>>s)
    	{
    		n1=0,n2=0,n3=0;
    		cin>>nb>>ns>>nc;
    		cin>>pb>>ps>>pc;
    		cin>>n;
    		for(int i=0;i<s.size();i++)
    		{
    			if(s[i]=='B')
    				n1++;
    			if(s[i]=='S')
    				n2++;
    			if(s[i]=='C')
    				n3++;
    		}
    		ll l=0,r=mmax;
    		ll ans,flag=0;
    		while(l<=r)
    		{
    			mid=(l+r)>>1;
    			if(getsum(mid)==n)
    			{
    				cout<<mid<<endl;
    				flag=1; 
    				break;
    			}
    			if(getsum(mid)<n)
    			{
    				l=mid+1;
    				ans=mid;
    			}
    			else
    			{
    				r=mid-1;
    			}
    		}
    		if(!flag)
    		cout<<ans<<endl;
    	 } 
    	return 0;
     }