牛客网暑期ACM多校训练营（第五场)
A-gpa
链接：https://www.nowcoder.com/acm/contest/143/A
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld
题目描述
Kanade selected n courses in the university. The academic credit of the i-th course is s[i] and the score of the i-th course is c[i].
At the university where she attended, the final score of her is
Now she can delete at most k courses and she want to know what the highest final score that can get.
输入描述:
The first line has two positive integers n,k
The second line has n positive integers s[i]
The third line has n positive integers c[i]输出描述:
Output the highest final score, your answer is correct if and only if the absolute error with the standard answer is no more than 10-5
示例1
输入
复制
3 1
1 2 3
3 2 1输出
复制
2.33333333333说明
Delete the third course and the final score is 备注:
1≤ n≤ 105
0≤ k < n
1≤ s[i],c[i] ≤ 103思路:二分查找/01分数规划
代码:
G-max
链接：https://www.nowcoder.com/acm/contest/143/G
来源：牛客网
Give two positive integer c, n. You need to find a pair of integer (a,b) satisfy 1<=a,b<=n and the greatest common division of a and b is c.And you need to maximize the product of a and b
题目意思:给一个数c,一个数n，从1-n中找两个数，这两个数数的最大公约数为c，求这两个数乘积的最大值。
链接：https://www.nowcoder.com/acm/contest/143/G
来源：牛客网
输入描述:
The first line has two positive integer c,n输出描述:
Output the maximum product of a and b.
If there are no such a and b, just output -1
示例1
输入
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2 4输出
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8说明
a=2,b=4备注:
1<=c,n<=10^9思路:看注释
代码:
# include
using namespace std;
typedef long long ll;
int main(){
ll c, n;
cin >> c >> n;
ll k1 = n/c, k2 = n/c - 1;
if(c > n)
cout << -1 << endl;
else if(c > n/2)//如果c大于n/2,那么找到的两个数的最大公约数只能c,并且那两个数也是c
cout << c*c << endl;
else if(n == c)//c==n的话就更显然了
cout << n*c << endl;
else{//这个就结论就厉害
k1 *= c, k2*= c;
cout << k1*k2 << endl;
}
return 0;
} J-plan
链接：https://www.nowcoder.com/acm/contest/143/J
来源：牛客网
题目描述
There are n students going to travel. And hotel has two types room:double room and triple room. The price of a double room is p2 and the price of a triple room is p3
Now you need to calulate the minimum total cost of these students.
输入描述:
The first line has three integers n, p2, p3输出描述:
Output the minimum total cost.
示例1
输入
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4 2 3输出
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4示例2
输入
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5 1 3输出
复制
3备注:
1<=n<=10^9
1<=p2,p3<=10^9思路:贪心，枚举就过了，不过要考虑清楚哦,注释上写了各种情况，似乎没考虑相除-1得负数的情况
代码:
#include
#include
#include
using namespace std;
typedef long long int ll;
int main()
{
ll p1,p2,p3,sum=0,yu;
double dp2,dp3;
cin>>p1>>p2>>p3;
dp2=double(p2)/2;
dp3=double(p3)/3;
if(dp2==dp3){//三人间，两人间单价相同，不浪费空间花费最低
if(p1%2) sum=(p1/2-1)*p2+p3;//余一个人
else sum=p1/2*p2;
cout<dp3){//三人间单价更低
yu=p1%3,sum=p1/3;//sum为开的间数
if(yu==1){//多一个人, // 2 3
//按四个人处理
int a,b,c;//直接比较三种情况
a=2*p2,b=p2+p3,c=p3+p3;
sum=(sum-1)*p3+min(a,min(b,c)); // 5 3 4
} // 2+3<3+3
else if(yu==2){//多两个人
if(p2 <
if(yu) {
if(p2*2>=p3){//开两间二人房大于等于开一间三人房的花费
sum=(sum-1)*p2+p3;
}
else sum=sum*p2+p2;
}
else sum=sum*p2;
cout<