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  • cf466A Cheap Travel

    A. Cheap Travel
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Ann has recently started commuting by subway. We know that a one ride subway ticket costs a rubles. Besides, Ann found out that she can buy a special ticket for m rides (she can buy it several times). It costs b rubles. Ann did the math; she will need to use subway n times. Help Ann, tell her what is the minimum sum of money she will have to spend to make n rides?

    Input

    The single line contains four space-separated integers nmab (1 ≤ n, m, a, b ≤ 1000) — the number of rides Ann has planned, the number of rides covered by the m ride ticket, the price of a one ride ticket and the price of an m ride ticket.

    Output

    Print a single integer — the minimum sum in rubles that Ann will need to spend.

    Sample test(s)
    input
    6 2 1 2
    
    output
    6
    
    input
    5 2 2 3
    
    output
    8
    
    Note

    In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three m ride tickets.

    题意是要买n张票,可以买1张价格是a,也可以买m张价格是b

    忘记考虑全买m张的最后剩下来直接买m张比一张一张更优的情况了……wa了两次

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<algorithm>
    #include<ctime>
    #define LL long long
    using namespace std;
    LL n,m,a,b;
    inline LL read()
    {
        LL x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    int main()
    {
    	n=read();m=read();a=read();b=read();
    	printf("%d",min(min(n*a,(n/m)*b+(n%m)*a),(n/m+1)*b));
    }
    

      

    ——by zhber,转载请注明来源
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  • 原文地址:https://www.cnblogs.com/zhber/p/4035996.html
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