POJ-3276（异或DP）

暴力方法的复杂度会是O(N^3)，而题目给定的N的范围是小于等于5000，暴力法必然会超时，考虑每一个位置的翻转情况，我们可以看到，其和前一个位置的翻转情况有密切联系：

pre[i] = post[i-k+1] ^ post[i-k+2] ^ ... ^ post[i-1]
post[i] = s[i] ^ pre[i]
====================================================
we have: pre[i-1] = post[i-k] ^ post[i-k+1] ^ ... ^ post[i-2]
thus: pre[i] = pre[i-1] ^ post[i-k] ^ post[i-1]

#include <stdio.h>

int N, K, M;
char s[5000] = {0};/* initial state, 'F' <=> 0, 'B' <=> 1 */
char pre[5000] = {0};/* whether i should be flipped by flips in front */
char post[5000] = {0};/* post[i] <=> whether to start a flip at i */

int main()
{
int i, k, m;
char c;
while(scanf("%d", &N) == 1){
K = 1;
M = 0;
for(i = 0; i < N; ++i){
scanf(" %c", &c);
if(s[i] = c != 'F') ++M;
}
for(k = 2; k <= N; ++k){
pre[0] = 0;
if(post[0] = s[0]) m = 1;
else m = 0;
for(i = 1; i + k <= N; ++i){
if(i >= k) pre[i] = pre[i-1] ^ post[i-k] ^ post[i-1];
else pre[i] = pre[i-1] ^ post[i-1];
if(post[i] = s[i] ^ pre[i]) ++m;
}
for(; i < N; ++i){
if(i >= k) pre[i] = pre[i-1] ^ post[i-k] ^ post[i-1];
else pre[i] = pre[i-1] ^ post[i-1];
if(pre[i] ^ s[i]) break;
post[i] = 0;
}
if(i == N && m < M) K = k, M = m;
}
printf("%d %d\n", K, M);
}
return 0;
}

由上面的状态方程我们可以看到pre[i]仅与pre[i-1]、post[i-k]、post[i-1]有关，我们完全可以省去pre[]数组而用一个滚动的变量来表示pre[i]：

#include <stdio.h>

int N, K, M;
char s[5000] = {0};/* initial state, 'F' <=> 0, 'B' <=> 1 */
char post[5000] = {0};/* post[i] <=> whether to start a flip at i */

int main()
{
int i, k, m;
char c, pre;
while(scanf("%d", &N) == 1){
K = 1;
M = 0;
for(i = 0; i < N; ++i){
scanf(" %c", &c);
if(s[i] = c != 'F') ++M;
}
for(k = 2; k <= N; ++k){
pre = 0;
if(post[0] = s[0]) m = 1;
else m = 0;
for(i = 1; i + k <= N; ++i){
if(i >= k) pre ^= post[i-k] ^ post[i-1];
else pre ^= post[i-1];
if(post[i] = s[i] ^ pre) ++m;
}
for(; i < N; ++i){
if(i >= k) pre ^= post[i-k] ^ post[i-1];
else pre ^= post[i-1];
if(s[i] ^ pre) break;
post[i] = 0;
}
if(i == N && m < M) K = k, M = m;
}
printf("%d %d\n", K, M);
}
return 0;
}