「CF466A」Cheap Travel
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 ntimes. Help Ann, tell her what is the minimum sum of money she will have to spend to make n rides?
The single line contains four space-separated integers n, m, a, b (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.
Print a single integer — the minimum sum in rubles that Ann will need to spend.
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1 |
6 2 1 2 |
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1 |
6 |
|
1 |
5 2 2 3 |
|
1 |
8 |
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.
题解
无语。。注意最后不满m也可以买一张m次的
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#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> #include<set> #include<queue> #include<map> #define pa pair<int,int> #define inf 1000000000 #define ll long long using namespace std; inline int read() { int 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 n,m,a,b; int main() { n=read(),m=read(),a=read(),b=read(); if((double)b/m>=a){printf("%d",a*n);return 0;} printf("%d",n/m*b+min(b,n%m*a)); return 0; } |
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