Codeforces Round #342 (Div. 2)(A)贪心,数学
2016-02-08 01:34
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A. Guest From the Past
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or
in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b)
rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984,
so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) —
the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) —
the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Sample test(s)
input
output
input
output
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
题意:你有n元钱,你有2种选择
1.使用每升a元买一瓶塑料牛奶
2使用每升b元买一瓶玻璃牛奶,每个玻璃瓶可以退回c元钱
问最多可以买多少升?
题解:如果b-c<=a,显然买b,那么只要n>=b,我们买b类绝对是不会亏的,我们尽可能的买b类后,再去买a类,否则直接买a类
我开始怎么都过不了样例,于是写了个DFS过了,最后极限数据还是超时了。
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <sstream>
#include <algorithm>
#define LL unsigned long long
using namespace std;
#define N 100000
#define inf 0x3f3f3f3f3f3f3f3f
LL n,a,b,c;
LL ans;
int main()
{
#ifdef CDZSC
freopen("i.txt","r",stdin);
#endif
while(~scanf("%lld",&n))
{
ans=0;
scanf("%lld%lld%lld",&a,&b,&c);
if(n>=b&&b-c<=a)
{
LL sum=((n-b)/(b-c)+1);
n-=sum*(b-c);
ans+=sum;
}
ans+=n/a;
printf("%lld\n",ans);
}
return 0;
}
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or
in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b)
rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984,
so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) —
the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) —
the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Sample test(s)
input
10 11 9 8
output
2
input
10 5 6 1
output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
题意:你有n元钱,你有2种选择
1.使用每升a元买一瓶塑料牛奶
2使用每升b元买一瓶玻璃牛奶,每个玻璃瓶可以退回c元钱
问最多可以买多少升?
题解:如果b-c<=a,显然买b,那么只要n>=b,我们买b类绝对是不会亏的,我们尽可能的买b类后,再去买a类,否则直接买a类
我开始怎么都过不了样例,于是写了个DFS过了,最后极限数据还是超时了。
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <sstream>
#include <algorithm>
#define LL unsigned long long
using namespace std;
#define N 100000
#define inf 0x3f3f3f3f3f3f3f3f
LL n,a,b,c;
LL ans;
int main()
{
#ifdef CDZSC
freopen("i.txt","r",stdin);
#endif
while(~scanf("%lld",&n))
{
ans=0;
scanf("%lld%lld%lld",&a,&b,&c);
if(n>=b&&b-c<=a)
{
LL sum=((n-b)/(b-c)+1);
n-=sum*(b-c);
ans+=sum;
}
ans+=n/a;
printf("%lld\n",ans);
}
return 0;
}
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