老方 • 1年前
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评论:
using namespace std; int a[100][100],n; int main(){
cin>>n;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>a[i][j];
}
}
for(int i=0;i<n;i++){
int maxn=-1;
int tmp;
for(int j=0;j<n;j++){
if(a[i][j]>maxn){
maxn=a[i][j];
tmp=j;
}
}
int minn=1001;
for(int k=0;k<n;k++){
if(a[k][tmp]<minn){
minn=a[k][tmp];
}
}
if(minn==maxn){
cout<<maxn;
}
}
return 0;
} 就这个
《爱与愁的故事第三弹·shopping》娱乐章。
调调口味来道水题。
爱与愁大神坐在公交车上无聊,于是玩起了手机。一款奇怪的游戏进入了爱与愁大神的眼帘:***(游戏名被打上了马赛克)。这个游戏类似象棋,但是只有黑白马各一匹,在点 x_1,y_1 和 x_2,y_2 上。它们得从点 x_1,y_1 和 x_2,y_2 走到 (1,1)。这个游戏与普通象棋不同的地方是:马可以走“日”,也可以像象走“田”。现在爱与愁大神想知道两匹马到 (1,1) 的最少步数,你能帮他解决这个问题么?
注意不能走到 x 或 y 坐标 \le 0 的位置。
第一行两个整数 x_1,y_1。
第二行两个整数 x_2,y_2。
第一行一个整数,表示黑马到 (1,1) 的步数。
第二行一个整数,表示白马到 (1,1) 的步数。
12 16
18 10
8
9
对于 100\% 数据,1\le x_1,y_1,x_2,y_2 \le 20。