A

#include <bits/stdc++.h>
using namespace std;
void solve()
{
    int n, m; 
    cin >> n >> m;
    vector<string> s(n);
    for(int i = 0; i < n; i++) cin >> s[i];
    for(int i = 0, j = 0; i < n; i++)
    {
        j += s[i].size();
        if(j > m)
        {
            cout << i << "\n";
            return;
        }
    }
    cout << n << "\n";
}

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

    int t; 
    cin >> t;
    while(t--) solve();

    return 0;
}

B
我们需要观察到一次操作只能改变奇偶性相隔的数 且需要观察奇偶平均数是否相等

# include <bits/stdc++.h>
using namespace std;
#define int long long 
const int N=2e5+10;
int a[N],n;

signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    int sum1=0,sum2=0,cnt1=0,cnt2=0;

    for(int i=1;i<=n;i++)
    {
    if(i&1)
    sum1+=a[i],cnt1++;
    else sum2+=a[i],cnt2++;    
    }
    if(sum1%cnt1||sum2%cnt2||sum1*cnt2!=sum2*cnt1)
    {
    cout<<"NO"<<endl;
    continue;
    }
    cout<<"YES"<<endl;
    }

}

C
可以被9整除的数各个数位上数和也为9的倍数 分类讨论即可

#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
    string s;
    cin>>s;
    int cnt2=0;
    int cnt3=0;
    int sum=0;
    for (int i=0;i<s.size();i++)
    {
        if (s[i]=='2')
            cnt2++;
        else if (s[i]=='3')
            cnt3++;
        sum+=(s[i]-'0');
    }
    int r=(9-sum%9);
    if (r==9)
    {
        cout<<"YES"<<"\n";
        return;
    }
    if (r&1)
        r+=9;
    r-=min(cnt3,(r/6))*6;
    int n=r/2;
    if (cnt2>=n)
        cout<<"YES"<<"\n";
    else
        cout<<"NO"<<"\n";
}
signed main()
{
    int T=1;
    cin>>T;
    while(T--)
    {
    solve();
    }
    return 0;
}

D

#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
    string s;
    cin>>s;
    for (int i = 0; i < (int)s.size(); i ++) 
    {
        int res = s[i], pos = i;
        for (int j = i + 1; j < min((int)s.size(), i + 11); j ++) 
            if ((int)s[j] - (j - i) > res) res = (int)s[j] - (j - i), pos = j;
        for (int j = pos; j > i; j --) swap(s[j], s[j - 1]), s[j - 1]--;
    }
    cout<<s<<endl;
}
signed main()
{
    int T=1;
    cin>>T;
    while(T--)
    {
        solve();
    }
    return 0;
}

F
根据同余可得a[i]-a[i+1]=(k[i+1]-k[i])*m,a[i]=k[i]m+b,k与m有关 即求数组差之间最大公约数 用线段树维护即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N =2e5+10;

int n, m;
LL w[N];
struct Node
{
    int l, r;
    LL sum, d,ma;
}tr[N * 4];

LL gcd(LL a, LL b)
{
    return b ? gcd(b, a % b) : a;
}

void pushup(Node &u, Node &l, Node &r)
{
    u.sum = l.sum + r.sum;
    u.ma=max(l.ma,r.ma);
    u.d = gcd(l.d, r.d);
}

void pushup(int u)
{
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}

void build(int u, int l, int r)
{
    if (l == r)
    {
        LL b = w[r] - w[r - 1];
        tr[u] = {l, r, b, b,w[l]};
    }
    else
    {
        tr[u].l = l, tr[u].r = r;
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}

void modify(int u, int x, LL v)
{
    if (tr[u].l == x && tr[u].r == x)
    {
        LL b = tr[u].sum + v;
        tr[u] = {x, x, b, b};
    }
    else
    {
        int mid = tr[u].l + tr[u].r >> 1;
        if (x <= mid) modify(u << 1, x, v);
        else modify(u << 1 | 1, x, v);
        pushup(u);
    }
}

Node query(int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r) return tr[u];
    else
    {
        int mid = tr[u].l + tr[u].r >> 1;
        if (r <= mid) return query(u << 1, l, r);
        else if (l > mid) return query(u << 1 | 1, l, r);
        else
        {
            auto left = query(u << 1, l, r);
            auto right = query(u << 1 | 1, l, r);
            Node res;
            pushup(res, left, right);
            return res;
        }
    }
}

int  query5(int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r) return tr[u].ma;
    int mid = tr[u].l + tr[u].r >> 1,maa=0;
    if (l<= mid) maa=query5(u << 1, l, r);
    if(r>mid)maa=max(maa,query5(u<<1+1,l,r));
    return maa;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
     int n,q;
     cin>>n>>q;
    for (int i = 1; i <= n; i ++ )cin>>w[i];
    for(int i=1;i<=n;i++)w[i]=w[i+1]-w[i];
    build(1, 1, n);
    int l, r;
    LL d;
    while (q-- )
    {
        cin>>l>>r;
        r--;
        if(r<l)cout<<0<<" ";
        else 
        {
        auto left = query(1, 1, l);
        Node right({0, 0, 0, 0});
        if (l + 1 <= r) right = query(1, l + 1, r);
        int x=abs(gcd(left.sum, right.d));
        //if(x==query5(1,l,r))x=0;
        cout<<x<<" ";
        }
    }
    cout<<'\n';
    }
    return 0;
}