1、http://codeforces.com/contest/408/problem/A
2、题目大意：
现在有n个队可以结账，每个队现在有ki个人，每个人有m件物品等待结账，每件物品结账5秒，每个人最终交钱需要15秒，求最快的结完账的队需要多长时间
题目很简单直接计算就行
3、题目：
Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are n cashiers at the exit from the supermarket. At the moment the queue for thei-th cashier already has ki people. The j-th person standing in the queue to the i-th cashier hasmi,?j items in the basket. Vasya knows that:
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.
The first line contains integer n (1?≤?n?≤?100) — the number of cashes in the shop. The second line containsn space-separated integers: k1,?k2,?...,?kn (1?≤?ki?≤?100), whereki is the number of people in the queue to thei-th cashier.
The i-th of the next n lines contains ki space-separated integers:mi,?1,?mi,?2,?...,?mi,?ki (1?≤?mi,?j?≤?100) — the number of products the j-th person in the queue for thei-th cash has.
Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.
1 1 1
20
4 1 4 3 2 100 1 2 2 3 1 9 1 7 8
100
In the second test sample, if Vasya goes to the first queue, he gets to the cashier in100·5?+?15?=?515 seconds. But if he chooses the second queue, he will need1·5?+?2·5?+?2·5?+?3·5?+?4·15?=?100 seconds. He will need1·5?+?9·5?+?1·5?+?3·15?=?100 seconds for the third one and7·5?+?8·5?+?2·15?=?105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue.
4、AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 0x7fffffff
int a[105];
int main()
{
int n,b,sum;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int minn=INF;
for(int i=1;i<=n;i++)
{
sum=0;
for(int j=1;j<=a[i];j++)
{
scanf("%d",&b);
sum+=b;
}
sum=sum*5+a[i]*15;
if(sum<minn)
minn=sum;
}
printf("%d\n",minn);
}
return 0;
}
Codeforces Round #239 (Div. 2) A. Line to Cashier(简单题),布布扣,bubuko.com
Codeforces Round #239 (Div. 2) A. Line to Cashier(简单题)
原文：http://blog.csdn.net/sdjzping/article/details/22594807