Go to movies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1029 Accepted Submission(s): 543
Problem Description
Winter holiday is coming!As the monitor, LeLe plans to go to the movies.
Because the winter holiday tickets are pretty expensive, LeLe decideds to try group-buying.
Because the winter holiday tickets are pretty expensive, LeLe decideds to try group-buying.
Input
There are multiple test cases, about 20 cases. The first line of input contains two integers n,m(1≤n,m≤100). n indicates the number of the students. m indicates how many cinemas have offered group-buying.
For the m lines,each line contains two integers ai,bi(1≤ai,bi≤100), indicating the choices of the group buying cinemas offered which means you can use bi yuan to buy ai tickets in this cinema.
For the m lines,each line contains two integers ai,bi(1≤ai,bi≤100), indicating the choices of the group buying cinemas offered which means you can use bi yuan to buy ai tickets in this cinema.
Output
For each case, please help LeLe **choose a cinema** which costs the least money. Output the total money LeLe should pay.
Sample Input
3 2
2 2
3 5
Sample Output
4
Hint
LeLe can buy four tickets with four yuan in cinema 1.Source
题意:在 m 个电影院选择一家进行团购,团购规则是 b 元买 a 张票,问买 n 张票所需的最少钱?
#include<iostream> #include<cstdio> #include<cstring> #include <algorithm> #include <math.h> using namespace std; typedef long long LL; int main() { int n,m; while(scanf("%d%d",&n,&m)!=EOF){ int MIN = 999999999; for(int i=1;i<=m;i++){ int a,b; scanf("%d%d",&a,&b); int k = n/a+((n%a==0)?0:1); MIN = min(MIN,b*k); } printf("%d ",MIN); } return 0; }