C

//观察数据得知是一维，发现性质非零和点的区间都可以。怎么dp，找每一次零和点，零和点间就是本次获得的
//并且前n次零和由dp【q】来，逆序方便处理

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

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);

    int t;
    cin >> t;
    while (t--) {
        int n;
        ll x;
        cin >> n >> x;
        vector<ll> a(n + 1), prefix(n + 1, 0);
        for (int i = 1; i <= n; ++i) {
            cin >> a[i];
            prefix[i] = prefix[i - 1] + a[i];
        }

        vector<ll> dp(n + 2, (ll)0);

        ll result = 0;

        for (int i = n - 1; i >= 0; --i) {
            int left = i + 1, right = n + 1;
            while (left < right) {
                int mid = left + (right - left) / 2;
                if (prefix[mid] - prefix[i] > x) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
            dp[i] = dp[left] + (left - i - 1);
            result += dp[i];
        }
        cout << result << '\n';
    }

    return 0;
}

C

//数据2e5状态这么多，一看就是线性
//分析去掉一个人后对前面人的影响，似乎没有，对后面人的影响，发现被迫选择的人可能会有影响
#include<bits/stdc++.h>
using namespace std;

void solve() {
    int n, m;
    cin >> n >> m;
    int nm = n + m + 1;

    vector<long long> a0(nm, 0), a1(nm, 0);
    for(int i = 0; i < nm; i++) cin >> a0[i];
    for(int i = 0; i < nm; i++) cin >> a1[i];

    vector<int> op(nm, 0);
    int bad = -1, badType = -1;
    int cur[2] = {0, 0};
    long long total_sum = 0;

    for(int i = 0; i < n + m; i++) {
        int curType = 0;
        if(a0[i] < a1[i]){
            curType = 1;
        }

        if(cur[curType] == ((curType == 0) ? n : m)){
            curType = 1 - curType;
            if(bad == -1){
                bad = i;
                badType = 1 - curType;
            }
        }

        op[i] = curType;
        if(curType == 0){
            total_sum += a0[i];
            cur[0]++;
        }
        else{
            total_sum += a1[i];
            cur[1]++;
        }
    }

    for(int i = 0; i < n + m; i++) {
        long long temp = total_sum - (op[i] == 0 ? a0[i] : a1[i]);

        if(bad != -1 && i < bad && op[i] == badType){
            if(badType == 0){
                temp += a0[bad] - a1[bad] + a1[nm-1];
            }
            else{
                temp += a1[bad] - a0[bad] + a0[nm-1];
            }
        }
        else{
            if(op[i] == 0){
                temp += a0[nm-1];
            }
            else{
                temp += a1[nm-1];
            }
        }

        cout << temp << " ";
    }
    cout << total_sum << "\n";
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}

B

//切道水题放松
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

void solve() {
    int n;
    cin >> n;

    vector<ll> a(n+1), b(n + 1);

    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }

    ll sum = 1;
    for (int i = 0; i < n; i++) {
        cin >> b[i];
        sum += abs(a[i] - b[i]);
    }
    cin >> b[n];
   // cout<<sum<<"\n";
    ll t=1e9;
    for (int i = 0; i < n; i++){
        ll mmax=max(a[i],b[i]);
        ll mmin=min(a[i],b[i]);
        if(b[n]<=mmax&&b[n]>=mmin){
            t=0;
            break;
        }else if(b[n]<mmin){
            t=min(t,mmin-b[n]);
        }else{
            t=min(t,b[n]-mmax);
        }

    }

    cout<<sum+t<<"\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

A

//暴力枚举不行！
//任敌杀戮不行。
//那就构造极限反例，k个字母全部要出现，取刚集满k的的字母放进ans，
//最后取一个缺的，不满足n个要随便输出点（忘记这个东西，一直卡住）
//思维路程
//我是尝试构造一个反例，比如对于k=3；就说abc，当n等于三的时候，
//最短的m应该是abcabcabc这样子才有全部子序列，但是对于题目给定的序列，
//我们就是要检查比如aabcabcbba，我们先检查字符串到哪里会集齐k然后模拟删去前缀
//同时把最后出现的字母放进答案，abcbba剩下，继续bba发现此时不符合，停止截断，
//然后我们发现只有第一个和第二个两个满足三个出现，
//但是n'等于3，错误，输出ans和缺的，如果仍不满足n就不断输出任意字母
#include <bits/stdc++.h>
using namespace std;

void solve() {
    int n, k, m;
    string s;
    cin >> n >> k >> m;
    cin >> s;

    int l = 0;
    int cnt_segments = 0; 
    map<char, int> freq; 
    vector<char> ans;
    while (l < m) {
        freq.clear();

        int r = l;
        while (r < m) {
            freq[s[r]]++;
            r++;
            if (freq.size() == k){ans.push_back(s[r-1]);break;}
        }

        if (freq.size() < k) break; 

        cnt_segments++;
        l = r;
    }

    if (cnt_segments >= n) {
        cout << "YES\n";
    } else {
        cout << "NO\n";
        for(auto c:ans){
            cout<<c;
        }
        freq.clear();
        for(int i=l;i<m;i++){
            freq[s[i]]++;
        }
        for(int i=0;i<k;i++){
            if(!freq['a'+i]){
                cout<<char('a'+i);
                break;
            }
        }
        while(ans.size()+1<n){
            cout<<'a';
            n--;
        }
        cout << "\n";
    }
}

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