(对结果分类讨论)牛客周赛 Round 1 C.游游的交换字符

题意：

思路：
观察发现相邻元素不同的结果只有两种，要么是101010101...要么是010101010，因此我们可以对结果分类讨论。直接模拟算出两种情况最少需要操作多少次，再取min即可。
需要注意的是，如果是奇数串，那么结果只有一种，数量多的一定要放两侧。

code:

#include <bits/stdc++.h>
#include<bits/extc++.h>
using namespace __gnu_pbds;
using namespace std;
using i64 = long long;
using u64 = unsigned long long;
using PII = pair<i64, i64>;
const int inf = 0x3f3f3f3f;
const i64 INF = 0x3f3f3f3f3f3f3f3f;
#define Z cout << "\n"
#define lb lower_bound
#define ub upper_bound
#define D(x) cerr << #x << ": " << (x) << "\n"
#define DV(v) cerr<<#v<<": ";for(int i=0;i<(v).size();i++)cerr<<((v)[i])<<",";cerr<<"\n"
#if 1
#define int i64
#endif
signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    string s;
    cin >> s;
    int n = s.size();
    s = ' ' + s;
    char c = '0';
    if (n & 1) {
        if (count(s.begin() + 1, s.end(), '0') < count(s.begin() + 1, s.end(), '1'))c = '1';
    }
    vector<int>p;
    p.push_back(5959);
    for (int i = 1; i <= n; i++) {
        if (s[i] == c)p.push_back(i);
    }
    int sum1 = 0;
    for (int i = 1; i < p.size(); i++) {
        int z = 2 * i;
        if (z <= n)sum1 += abs(z - p[i]);
        else sum1 = inf;
    }
    int sum2 = 0;
    for (int i = 1; i < p.size(); i++) {
        int z = 2 * i - 1;
        if (z <= n)sum2 += abs(z - p[i]);
        else sum2 = inf;
    }
    cout << min(sum1, sum2);
    return 0;
}