本篇题解只是记录我的做题过程代码，不提供最优解 （另外这里所有的时间复杂度都只分析单个样例，不算 t t t的时间复杂度）

A

. 本题设计算法：模拟

时间复杂度 O ( n ) O(n) O(n)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10,INF = 1e9;
int a[N];
map<int,int> st;

void solve()
{
    int n;
    cin >> n;
    for (int i = 1;i <= n;i ++ ) cin >> a[i];

    int x = 1,first = -1,second = -1;
    for (int i = 1;i <= n;i ++ ) {
        if (a[i] == 0 && first == -1) {
            first = i;
            for (int j = i;j <= n;j ++ ) {
                if (a[j] == 1) {
                    i = j - 1;
                    break;
                }else x ++;
            }

        } else if (a[i] == 0 && first != -1) {
            second = i;
            break;
        }
        //cout << i <<   << x <<
;
    }

    int index = -1;
    for (int i = n;i >= 1;i -- ) {
        if (a[i] == 0 ) {
            index = i;
            break;
        }
    }

    if (first == -1) cout << 0 <<
;
    else if (second == -1) cout << x << 
;
    else cout << index - first + 2 << 
;
}
int main()
{
    int t;
    cin >> t;
    while (t -- ) {
        solve();
    }
    return 0;
}

B

. 本题设计算法：贪心 贪心策略：只要找到传球数最大的球员与剩余的球员对比即可

时间复杂度 O ( n ) O(n) O(n)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
typedef long long ll;
const ll N = 2e5 + 10,INF = 1e9 + 7;
ll a[N];

void solve()
{
    ll n,maxn = -INF;
    cin >> n;
    ll sum = 0;
    for (int i = 1;i <= n;i ++ ) {
        cin >> a[i];
        maxn = max(maxn,a[i]);
        sum += a[i];
    }

    ll yu = sum - maxn;
    if (maxn == 0) cout << 0 << 
;
    else if (maxn <= yu) cout << 1 <<
;
    else {
        yu ++;
        ll res = maxn - yu;
        cout << res + 1 << 
;
    }
}
int main()
{
    int t;
    cin >> t;
    while (t -- ) {
        solve();
    }
    return 0;
}

C

. **本题设计算法：关键字排序 + 快速组合 **

先分别对每种颜色的所有可能坐标进行统计

然后枚举所有颜色，求曼哈顿距离的和【注意对x与y坐标排序（这样可以忽略曼哈顿距离的绝对值，从而比较好求），这里分开x，y坐标求比较好】

另外注意，同颜色求和是要找到 C l e n 2 C_{len}^2 Clen2组合相减的规律，用O(n)的时间求出，而不是暴力求，会超时。

举个栗子： 1 2 3 4 的 C 4 2 C_4^2 C42的组合相减

时间复杂度 O ( n 2 ∗ l o g n ) O(n^2 * logn) O(n2∗logn)

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
const int N = 2e5 + 10;
typedef long long ll;
typedef pair<ll,ll> PII;
vector<PII> ver[N];
ll res,d;

bool cmp(PII x,PII y)
{
    return x.second < y.second;
}
int main()
{
    ll n,m;
    cin >> n >> m;
    for (ll i = 1;i <= n;i ++ ) {
        for (ll j = 1;j <= m;j ++ ) {
            ll val;
            cin >> val;
            ver[val].push_back({i,j});
        }
    }

    for (ll i = 1;i <= N;i ++ ) {
        ll len = (ll)ver[i].size();
        if (len <= 1) continue;

        sort (ver[i].begin(),ver[i].end());
        for (ll j = 1;j < len;j ++ ) res += (len - j) * j * (ver[i][j].first - ver[i][j - 1].first);//优化重点
		
        sort (ver[i].begin(),ver[i].end(),cmp);
        for (ll j = 1;j < len;j ++ ) res += (len - j) * j * (ver[i][j].second - ver[i][j - 1].second);//优化重点
    }

    cout << res << 
;
    return 0;
}