Codeforces--616B--Dinner with Emma（模拟水题）

Dinner with Emma

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

Jack decides to invite Emma out for a dinner. Jack is a modest student, he doesn't want to go to an expensive restaurant. Emma is a girl with high taste, she prefers elite places.

Munhattan consists of n streets and m avenues. There is exactly one restaurant on the intersection of each street and avenue.
The streets are numbered with integers from 1 to n and the avenues are numbered with integers from 1 to m.
The cost of dinner in the restaurant at the intersection of the i-th street and the j-th avenue is cij.

Jack and Emma decide to choose the restaurant in the following way. Firstly Emma chooses the street to dinner and then Jack chooses the avenue. Emma and Jack makes their choice optimally: Emma wants to maximize the cost of the dinner, Jack wants to minimize
it. Emma takes into account that Jack wants to minimize the cost of the dinner. Find the cost of the dinner for the couple in love.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 100) — the number of streets and avenues in Munhattan.

Each of the next n lines contains m integers cij (1 ≤ cij ≤ 109)
— the cost of the dinner in the restaurant on the intersection of the i-th street and the j-th avenue.

Output

Print the only integer a — the cost of the dinner for Jack and Emma.

Sample Input

Input

3 4
4 1 3 5
2 2 2 2
5 4 5 1

Output

2

Input

3 3
1 2 3
2 3 13 1 2

Output

1

Hint

In the first example if Emma chooses the first or the third streets Jack can choose an avenue with the cost of the dinner 1. So she chooses the second street and Jack chooses any avenue. The cost of
the dinner is 2.

In the second example regardless of Emma's choice Jack can choose a restaurant with the cost of the dinner 1.

Source

Educational Codeforces Round 5

题意：两个人去吃饭一个人A想吃贵的，一个人B想吃便宜的，现在有n条街，每条街m个餐厅，现在给出所有的餐厅消费情况，先由A挑选在哪一条街吃饭，然后B选餐厅。

找每条街的最小值然后从最小值中取最大，数据可能有点大，选取最小值的时候minn要定义的大一些，因为少一个3f，哇了四次

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
__int64 a;
__int64 maxx=0,minn;
for(int i=0;i<n;i++)
{
minn=0x3f3f3f3f;
for(int j=0;j<m;j++)
{
scanf("%I64d",&a);
minn=min(a,minn);
}
maxx=max(minn,maxx);
}
printf("%I64d\n",maxx);
}
return 0;
}