poj 3276 开关反转

分析：

通过性质分析，把本来很复杂的问题简化了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define pr(x) cout << #x << ": " << x << "  " 
#define pl(x) cout << #x << ": " << x << endl;
const int maxn = 5000 + 13;

struct jibancanyang
{
    int n;
    bool G[maxn], f[maxn]; 

    int getTimes(int k) {
        int sum = 0, ret = 0;
        for (int i = 0; i < n; ++i) {
            if (i - k >= 0) sum -= f[i - k];
            if (i <= n - k) {
                if (sum % 2) {
                    if (!G[i]) f[i] = 0; 
                    else f[i] = 1, ret++;
                } else {
                    if (G[i]) f[i] = 0;
                    else f[i] = 1, ret++;
                }
            } else {
                f[i] = 0;
                if (!G[i] && sum % 2 == 0) return maxn;
                if (G[i] && sum % 2) return maxn;
            }
            sum += f[i];
        }
        return ret;
    }

    void fun() {
        while (cin >> n) {
            for (int i = 0; i < n; ++i) {
                char c;
                cin >> c;
                G[i] = c == 'B' ? 0 : 1;
            }
            int ans = maxn, K;
            for (int k = 1; k <= n; ++k) {
                int temp = getTimes(k);
                if (temp < ans) ans = temp, K = k;
            }
            cout << K << " " << ans << endl;
        }
    }

}ac;

int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif
    ac.fun();
    return 0;
}