[比赛链接](Dashboard - Educational Codeforces Round 155 (Rated for Div. 2) - Codeforces)

A. Rigged!

思路：如果存在 力量和耐力同时 >= 第一位选手的则输出-1， 否则直接输出第一位选手的力量即可 

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

void solve() {
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for(int i = 0; i < n; i++) cin >> a[i] >> b[i];
    for(int i = 1; i < n; i++) {
        if(a[i] >= a[0] && b[i] >= b[0]) {
            cout << "-1\n";
            return;
        }
    }
    cout << a[0] << endl;
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

B. Chips on the Board

思路： 结果，每一行都存在一个芯片或者每一列都存在一个芯片。 假设某一列没有芯片，那一定每一行都有一个芯片，

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

void solve() {
    int n;
    cin >> n;
    ll sum1 = 0, sum2 = 0;
    vector<int> a(n), b(n);
    for(int i = 0; i < n; i++) cin >> a[i], sum1 += a[i];
    for(int i = 0; i < n; i++) cin >> b[i], sum2 += b[i];
    ll t1 = *min_element(a.begin(), a.end());
    ll t2 = *min_element(b.begin(), b.end());
    cout << min(sum1 + n * t2, sum2 + n * t1) << endl;
}

int main() {
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

C. Make it Alternating

思路：组合数学， 一段连续（长度为x）相同的字符必须删掉x - 1个， 删除的方法就是comb(len, len - 1).

再计算一下所有要删除的个数 cnt， fac[cnt] 乘进去即可。

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#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
typedef long long ll;
const int N = 2e5 + 10;
ll qui(ll a, ll b)
{
    ll res = 1;
    while (b)
    {
        if (b & 1)
            res = res * a % MOD;
        b >>= 1;
        a = a * a % MOD;
    }
    return res;
}

ll fac[N], inv_fac[N];

void init()
{
    fac[0] = 1;
    for (int i = 1; i < N; i++)
        fac[i] = fac[i - 1] * i % MOD;
    inv_fac[N - 1] = qui(fac[N - 1], MOD - 2);
    for (int i = N - 1; i > 0; i--)
        inv_fac[i - 1] = inv_fac[i] * i % MOD;
}

ll comb(int a, int b)
{
    return fac[a] * inv_fac[b] % MOD * inv_fac[a - b] % MOD;
}

void solve() {
    string s;
    cin >> s;
    vector<int> a;
    int cnt = 0;
    for(int i = 0; i < s.size(); i++) {
        if(i && s[i] != s[i - 1]) {
            a.push_back(cnt);
            cnt = 1;
        }else cnt++;
    }
    if(cnt) a.push_back(cnt);
    cout << s.size() - a.size() << ' ';
    vector<int> b;
    cnt = 0;
    for(auto x : a) if(x > 1) b.push_back(x), cnt += (x - 1);
    ll ans = 1;
    for(auto x : b) {
        ans = (ans * comb(x, x - 1)) % MOD; 
    }
    ans = (ans * fac[cnt]) % MOD;
    cout << ans << endl;
}

int main() {
    init();
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}

D. Sum of XOR Functions

思路 : 贡献法， 拆分成二进制的，计算每一位的贡献。 

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MOD = 998244353;

void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    ll ans = 0;
    for(int b = 0; b < 30; b++) {
        vector<ll> cnt(2);
        vector<ll> sumO(2);
        int x = 0;
        ll cur = 0;
        cnt[0] = 1;
        for(int i = 0; i < n; i++) {
            x ^= ((a[i] >> b) & 1);
            ll sum = (cnt[x ^ 1] * (i + 1)) % MOD;
            cur = (cur + ((sum - sumO[x ^ 1]) + MOD) % MOD) % MOD;
            cnt[x]++;
            sumO[x] = (sumO[x] + (i + 1)) % MOD;
        }
        ans = (ans + (1 << b) * cur) % MOD;
    } 
    cout << ans << endl;
}


int main() {
    solve();
    return 0;
}