CF- Educational Codeforces Round 97 (Rated for Div. 2)-1437B. Reverse Binary Strings【思维/贪心】

题目链接
题意：给定一个01数量相等的二进制字符串，每次操作可以旋转一段子串，问至少多少次操作可以使得所有相邻字符都不一样。
思路：对于某个"00"开头（结尾），"11"结尾（开头）的子串，可以在一次操作完成置换，所以答案就是字符串中"00"或者"11"的数量的最大值，将其与整个字符串的开头或结尾配对即可。
AC代码：

#include <bits/stdc++.h>
#define pcc pair<char, char>
#define pii pair<int, int>
#define vi vector<int>
#define vl vector<ll>
#define rep(i, x, y) for (int i = x; i < y; i++)
#define per(i, x, y) for (int i = x; i >= y; i--)
#define rep0(i, n) for (int i = 0; i < (n); i++)
#define per0(i, n) for (int i = (n)-1; i >= 0; i--)
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define sz(x) (x).size()
#define all(x) (x).begin(), (x).end()
#define ll long long
#define ull unsigned long long
#define db double
#define ld long double
using namespace std;inline int read() {int x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-') f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0';ch = getchar();}return x * f;
}const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 6e6 + 10;
const int MAX = 1e5;
const double pi = acos(-1.0);
const ll INF = 0x3f3f3f3f3f3f3f;
ll qpow(ll a, ll b) {ll res = 1;while (b) {if (b & 1) res = (res * a) % mod;a = (a * a) % mod;b >>= 1;}return res % mod;
}
/***************main****************/
ll T = 1;
ll m, n;
int main() {T = read();while (T--) {n = read();int ans[2] = {0, 0};string s;cin >> s;rep(i, 1, n) if (s[i] == s[i - 1]) ans[s[i] - '0']++;cout << max(ans[0], ans[1]) << endl;}return 0;
}

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CF- Educational Codeforces Round 97 (Rated for Div. 2)-1437B. Reverse Binary Strings【思维/贪心】

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