题意: 有n*n的方格棋盘,现有黄蓝两种颜色,一圈一圈的图色,使得上一圈与下一圈相邻。试问总共最少要图多少钱?
思路: 嗐,太菜了,被这个题卡了好久,还是队友先写出来。 答案就是n/2+1。
代码实现:
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int t, n;
signed main()
{
IOS;
cin >> t;
while(t --){
cin >> n;
cout << n/2 + 1 << endl;
}
return 0;
}
来源:oschina
链接:https://my.oschina.net/u/4350320/blog/4484750