E. 2584
原题链接
思路
- 需要注意最后的答案会超过 2584,所以需要多算几项斐波那契数列
小技巧:只需要写一个向左移的函数,别的方向可以先旋转/翻转再向右移然后再转回来
代码
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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
#define X first
#define Y second
typedef pair<int,int> pii;
typedef long long LL;
const char nl = '\n';
const int N = 110;
const int M = 2e5+10;
int n,m;
LL a[N][N],b[N][N];
LL f[N];
bool cnt[N];
int find(LL x){ //二分搜索该数在斐波那契数列中的下标(离散化)
int l = 1, r = 90;
while(l < r){
int mid = l + r >> 1;
if(f[mid] >= x)r = mid;
else l = mid + 1;
}
return l;
}
void init(){
f[1] = 1,f[2] = 2;
for(int i = 3; i <= 90; i ++ )f[i] = f[i - 1] + f[i - 2]; //92以上爆longlong
}
void rotate_90(){ //顺时针旋转90°
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= n; j ++ )
b[i][j] = a[n+1-j][i];
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= n; j ++ )
a[i][j] = b[i][j];
}
void rotate_180(){ //旋转180°
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= n; j ++ )
b[i][j] = a[n+1-i][n+1-j];
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= n; j ++ )
a[i][j] = b[i][j];
}
void rotate_270(){ //逆时针旋转90°
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= n; j ++ )
b[i][j] = a[j][n+1-i];
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= n; j ++ )
a[i][j] = b[i][j];
}
void work(){
for(int i = 1; i <= n; i ++ ){
memset(cnt,0,sizeof cnt); //使用情况记录范围以行为单位
for(int j = 2; j <= n; j ++ ){ //1前面没有数,因此可以从2开始
if(a[i][j] == -1 || !a[i][j])continue; //只有正整数才会移动
int t = j; //记录该数的位置(不然会改变j造成死循环)
for(int k = j - 1; k >= 1; k --){
if(a[i][k] == -1)break; //遇到障碍停止
else if(!a[i][k]){
swap(a[i][k],a[i][t]); //交换位置(即为前进)
t = k; //注意要更新实时位置
}
else{
int l = find(a[i][k]), r = find(a[i][t]); //找到在斐波那契数列中的为位置
if(abs(l-r)==1 && !cnt[k]){ //如果位置相邻且当前所在下标未发生改变
a[i][k]+=a[i][t]; //求和
a[i][t]=0; //取0
cnt[k] = 1; //记录
}
break; //遇到正整数停止
}
}
}
}
}
void out(){
for(int i = 1; i <= n; i ++ ){
for(int j = 1; j <= n; j ++ ){
cout << a[i][j] << " ";
}
cout << nl;
}
}
void solve(){
init();
cin >> n >> m;
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= n; j ++ )
cin >> a[i][j];
string q;
cin >> q;
for(auto x:q){ //q次转换
if(x == 'L'){
work();
}
else if(x == 'R'){
rotate_180(); //旋转180°(right->left)
work();
rotate_180();
}
else if(x == 'U'){
rotate_270(); //逆时针旋转90°(up->left)
work();
rotate_90(); //顺时针旋转90°(恢复旋转)
}
else{
rotate_90(); //顺时针旋转90°(down->left)
work();
rotate_270(); //逆时针旋转90°(恢复旋转)
}
}
out();
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
solve();
}
拓展
矩阵的变幻