https://codeforces.com/contest/1180
A. Alex and a Rhombus
找规律，不难发现 a[i] = a[i-1]+4*(i-1)
或者 a[n] = 2*n*n-2*n+1
#include<bits/stdc++.h> using namespace std; int main(){ int n,t,a[200]; cin>>n; a[1] = 1; a[2] = 5; for(int i = 3; i <= n; i++) a[i] = a[i-1]+4*(i-1); cout<< a[n] <<endl; return 0; }
B. Nick and Array
给定一个序列，可以对其中的值取负-1， 问乘积最大且操作次数最小的序列是（若多种输出一种）
当a_i > 0时，变为 -a_i-1 |product |会增大，所以当product 为正值时，已经找到答案
当product < 0 时（即n&1），把最大的负数变为正数就可以，证明的话随手写三个负数 纸上乘下一看即知
#include<bits/stdc++.h> using namespace std; const int MA = 1e5+100; //int a[MA],b[MA],c[MA]; vector<int> a; int MI = 1e6+100; int MII = -1e6; int main(){ int n,t1 = 0, t2 = 0; long long sum = 1; scanf("%d", &n); int cnt, tmp, MII = 0; for(int i = 0; i < n; i++) { int k; scanf("%d", &k); a.push_back(k); if(a[i] >= 0) a[i] = -a[i]-1; if(a[i] <= MII ) tmp = i, MII = a[i]; } if(n&1) a[tmp] = -a[tmp]-1; { for(int i = 0; i < n; i++) cout<<a[i] <<" ";} return 0; }
C. Valeriy and Deque
enmmmmm 这道题教我用上了deque，观察以后发现当A变成序列里最大的数以后就不会变化了，也就是A固定 剩下的数成一个序列左移成B
在A不是最大的数之前，扫一遍O(n)，将所有情况用vector<pair<int, int> >v记录下来， 当操作数 m_j > n 时，B_j = v[(m_j - maxIndex-1)%(n-1)+maxIndex].second
/* 排首 两个A B， 若A > B, A~~~~B 否 B~~~~A input 5 3 1 2 3 4 5 1 2 10 output 1 2 2 3 5 2 input 2 0 0 0 output */ #include<algorithm> #include<iostream> #include<vector> #include<deque> using namespace std; int main(){ int n,m; cin>>n>>m; deque<int> num; int maxValue = 0; for(int i = 0; i < n; i++) { int x;cin>>x; num.push_back(x); if(x > maxValue) maxValue = x; } vector<pair<int, int> > results; while(num.front() != maxValue) { int a = num.front(); num.pop_front(); int b = num.front(); num.pop_front(); results.push_back(make_pair(a, b)); if(a < b) swap(a, b); num.push_back(b); num.push_front(a); } int len = results.size(); num.pop_front(); for(int i = 0; i < n-1; i++) { int x = num.front(); results.push_back(make_pair(maxValue, x)); num.pop_front(); } for(int i = 0; i < m; i++){ long long k; cin>>k; if(k <= n) cout<<results[k-1].first <<" "<<results[k-1].second <<endl; else{ int x = (k-len-1)%(n-1) + len; cout<<results[x].first <<" " <<results[x].second << endl; } } return 0; }
D. Tolik and His Uncle
输入n，m （1<= n*m <= 1e6），代表一个n*m的网格 左上角坐标1，1 右下角坐标 n，m
enmmmmmm 然后从1，1 开始遍历（可以到任意网格中的任意点），要求每个点都要走过，且两两之间的向量不能相同，
例如 1,1 -> 1,2 以后 1,2 -> 1,3就不行 1,1 -> 2,2 以后 2,2 -> 3,3就不行
First, we are going to describe how to bypass 1⋅m1⋅m strip.
This algorithm is pretty easy — (1,1)(1,1) -> (1,m)(1,m) -> (1,2)(1,2) -> (1,m−1)(1,m−1) -> ……. Obviously all jumps have different vectors because their lengths are different.
It turns out that the algorithm for n⋅mn⋅m grid is almost the same. Initially, we are going to bypass two uttermost horizontals almost the same way as above — (1,1)(1,1) -> (n,m)(n,m) -> (1,2)(1,2) -> (n,m−1)(n,m−1) -> …… -> (1,m)(1,m) -> (n,1)(n,1). One can realize that all vectors are different because they have different dydy. Note that all of them have |dx|=n−1|dx|=n−1. Then we will jump to (2,1)(2,1) (using (−(n−2),0)(−(n−2),0) vector). Now we have a smaller task for (n−2)⋅m(n−2)⋅m grid. One can see that we used only vectors with |dx|≥n−2|dx|≥n−2, so they don't influence now at all. So the task is fully brought down to a smaller one.
意思是从1，1左上角开始 一个右下角n，m 再左上角 1，2 再右下角n，m-1 这样就可以保证所有的vector都不同，但是要分下n的奇偶，因为若n为奇数，则n/2+1 这一行需要单独讨论，
n为奇数的话就可以从 x，1->x，m -> x,2 -> x,m-1 .........(x=n/2+1) ，若m为基数则再输出一个 n/2+1，m/2+1
开始没想到，看题解以后 用了双端队列 TLE on test 6，一直TLE 很是不解， 直到想起输入能卡人 输出也能？
enmmmmmm 答案是能， 把endl 改成 就过了
#include<algorithm> #include<iostream> #include<deque> using namespace std; int main(){ ios_base::sync_with_stdio(false); cin.tie(0); cout.precision(20); cout << fixed; int n,m; cin>>n >>m; for(int i = 0; i < n/2; i++) for(int j = 0; j < m; j++) { cout<<i+1 <<" "<<j+1<<" "; cout<<n-i <<" "<<m-j<<" "; } if(n & 1) { int x = n/2 + 1; cout<< x<<" 1 "; deque<int > num; for(int i = 2; i <= m; i++) num.push_back(i); while(!num.empty()){//!num.empty() cout<<x <<" " <<num.back() <<" "; num.pop_back(); if(!num.empty()) {//!num.empty() cout<<x <<" " <<num.front()<<" "; num.pop_front(); } } } return 0; } /* inputCopy 2 3 outputCopy 1 1 1 3 1 2 2 2 2 3 2 1 inputCopy 1 1 outputCopy 1 1 */
直接输出，不用双端队列也行， 而且更节省空间
#include<cstdio> #include<algorithm> #include<cstring> using namespace std; int n,m; int main(){ scanf("%d%d",&n,&m); for(int i=1;i<=n/2;i++){ for(int j=1;j<=m;j++){ printf("%d %d ",i,j); printf("%d %d ",n-i+1,m-j+1);//将 } } if(n&1){ for(int i=1;i<=m/2;i++){ printf("%d %d ",n/2+1,i); printf("%d %d ",n/2+1,m-i+1); } if(m&1)printf("%d %d ",n/2+1,m/2+1); } return 0; }
764ms,300KB -> 234ms,0KB