P2882 [USACO07MAR]Face The Right Way G POJ3276

有个01串，每次可以将k个01串逆转，求最少操作次数的情况下的最小k
区间修改用差分实现

#include<bits/stdc++.h>
#define rep(i,j,k) for(int i(j);i<=k;++i)
#define drp(i,j,k) for(int i(j);i>=k;--i)
#define repg(x) for(int i(G.head[x]);i;i=G.next[i])
#define bug cout<<"~~~~~~~~~~~~~"<<'
';
using std::cin;
using std::cout;
const int N = 5e3 + 79;
int n;
int a[N], b[N], sum, tag, ans=N,anss;
char op;
int main() {
	cin >> n;
	rep(i, 1, n) {
		cin >> op;
		a[i] = (op == 'F')  ;
	}
	rep(k, 1, n) {
		int now=0 , tot = 0;
		int flag = 1;
		memset(b, 0, sizeof b);
		rep(i, 1, n) {
			now ^= b[i];
			if(a[i]^now == 0) {
				if(i + k - 1 > n) {
					flag = 0;
					break;
				}
				tot++;
				now^=1;
				b[i+k]^=1; 
			}
		}
		if(flag){
			if(tot<ans){
				ans=tot;
				anss=k;
			}
		}
	}
	cout<<anss<<' '<<ans<<'
';
	return 0;
}