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codeforces 484C Strange Sorting Codeforces Round #276 (Div. 1) C

思路:首先 他是对1到k 元素做一次变换,然后对2到k+1个元素做一次变化。。。。依次做完。

       如果我们对1到k个元素做完一次变换后,把整个数组循环左移一个。那么第二次还是对1 到 k个元素做和第一次一样的变换,再左移,再对1 到 k个元素做和第一次一样的变换,依次做完n-k+1即可。

       假设题目要求的变换为C    循环左移变换为P。那么对于每次查询 相当于做  n-k+1  (CP) 变换。最后把答案再向右移动n-k+1  回到原来位置即可。

       那么问题就解决了   效率    每次查询n log(n-k+1)   

 

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#include<string>#include <iostream>#include<vector>#define MAXN 1000050#define lson i<<1#define rson i<<1|1#define LL long longusing namespace std;char s[MAXN];char a[MAXN];int p[MAXN];int ans[MAXN];int tmp[MAXN];int main() {	int n, m;	scanf(" %s", s);	n = strlen(s);	scanf("%d", &m);	for (int i = 0; i < m; ++i) {		int k, d;		scanf("%d%d", &k, &d);		for (int i = 0; i < n; ++i)			p[i] = ans[i] = i;		int cid = 0;		for (int x = 0; x < d; ++x)			for (int j = x; j < k; j += d)			{				p[cid++] = j;			}		int last = p[0];		for (int j = 0; j <n-1;++j)			p[j] = p[j + 1];		p[n-1] = last;		int x = n - k + 1;		while (x) {			if (x & 1) {				for (int j = 0; j < n; ++j)					tmp[j] = ans[p[j]];				for(int j=0;j<n;++j)					ans[j]=tmp[j];			}			for (int j = 0; j < n; ++j)				tmp[j] = p[p[j]];			for (int j = 0; j < n; ++j)				p[j] = tmp[j];			x >>= 1;		}		for (int j = 0; j < n; ++j)		{			a[j] = s[ans[(j + k - 1) % n]];		}		for (int j = 0; j < n; ++j)			s[j] = a[j];		printf("%s\n", s);	}}

  

codeforces 484C Strange Sorting Codeforces Round #276 (Div. 1) C