No clue as to what's wrong. Can anyone please give me a hint or what's the case?

n = 228 -> ans = 2509198527
n = 121 -> ans = 2368799
n = 10 -> ans = 9
n = 150 -> ans = 19406015
n = 201 -> ans = 517361669
n = 300 -> ans = 114872472063
n = 500 -> ans = 732986521245023

I have WA8. Who can help with test?

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;
struct dw
{
    ll a,f,s;
}A[1600];
bool cmp(dw n1,dw n2)
{
    if(n1.a==n2.a)
    {
        if(n1.f==n2.f)return n1.s<n2.s;
        else return n1.f<n2.f;
    }
    else return n1.a<n2.a;
}
ll ys1[1600],ys2[1600];
ll ans,ad,au,al,ar;
struct node
{
    ll l,r,lc,rc,lmax,ln,rmax,rn,mmax,sum,nn;
    ll L,R,lR,rL;
    bool have;
}tr[3100];ll trlen=0;
ll da=1501;
void update(ll now)
{
    ll lc=tr[now].lc,rc=tr[now].rc;
    tr[now].sum=tr[lc].sum+tr[rc].sum;
    tr[now].nn=tr[lc].nn+tr[rc].nn;
    ll qwq=-da,ql,qr;
    if(tr[lc].have){if(tr[lc].mmax>qwq){qwq=tr[lc].mmax;ql=tr[lc].L,qr=tr[lc].R;}}
    if(tr[rc].have){if(tr[rc].mmax>qwq){qwq=tr[rc].mmax;ql=tr[rc].L,qr=tr[rc].R;}}
    if(tr[lc].rn+tr[rc].ln>=2){if(tr[lc].rmax+tr[rc].lmax>qwq){qwq=tr[lc].rmax+tr[rc].lmax;ql=tr[lc].rL;qr=tr[rc].lR;}}
    if(qwq==-da)tr[now].have=false,tr[now].mmax=tr[now].L=tr[now].R=0;
    else tr[now].have=true,tr[now].mmax=qwq,tr[now].L=ql,tr[now].R=qr;
    tr[now].lmax=tr[lc].lmax,tr[now].ln=tr[lc].ln,tr[now].lR=tr[lc].lR;
    if(tr[lc].sum+tr[rc].lmax>=tr[now].lmax)tr[now].lmax=tr[lc].sum+tr[rc].lmax,tr[now].ln=tr[lc].nn+tr[rc].ln,tr[now].lR=tr[rc].lR;
    tr[now].rmax=tr[rc].rmax,tr[now].rn=tr[rc].rn,tr[now].rL=tr[rc].rL;
    if(tr[rc].sum+tr[lc].rmax>=tr[now].rmax)tr[now].rmax=tr[rc].sum+tr[lc].rmax,tr[now].rn=tr[rc].nn+tr[lc].rn,tr[now].rL=tr[lc].rL;
}
void bt(ll l,ll r)
{
    ll now=++trlen;tr[now].l=l;tr[now].r=r;tr[now].lc=tr[now].rc=-1;
    tr[now].mmax=0;tr[now].have=0;tr[now].lmax=tr[now].rmax=0;tr[now].ln=tr[now].rn=0;tr[now].sum=tr[now].nn=0;
    tr[now].L=tr[now].R=tr[now].lR=tr[now].rL=0;
    if(l<r)
    {
        ll mid=(l+r)/2;
        tr[now].lc=trlen+1;bt(l,mid);
        tr[now].rc=trlen+1;bt(mid+1,r);
        update(now);
    }
    else tr[now].lR=tr[now].rL=l;
}
void change(ll now,ll x,ll k)
{
    if(tr[now].l==tr[now].r)
    {
        tr[now].sum+=k;tr[now].nn++;
        tr[now].lmax=tr[now].rmax=tr[now].sum;
        tr[now].ln=tr[now].rn=tr[now].nn;
        if(tr[now].nn>=2)tr[now].mmax=tr[now].sum,tr[now].L=tr[now].R=x,tr[now].have=true;
    }
    else
    {
        ll lc=tr[now].lc,rc=tr[now].rc,mid=(tr[now].l+tr[now].r)/2;
        if(x<=mid)change(lc,x,k);
        else change(rc,x,k);
        update(now);
    }
}
ll n,z1,z2;
void solve()
{
    for(ll i=1;i<=n;i++)
    {
        trlen=0;bt(1,z2);
        ll st1=A[i].a,st=i;
        for(ll j=st1;j<=z1;j++)
        {
            bool bk=false;
            while(st<=n&&A[st].a==j){change(1,A[st].f,A[st].s);st++;bk=1;}
            if(bk)
            {
                if(tr[1].have)
                {
                    if(ans<tr[1].mmax)
                    {
                        ans=tr[1].mmax;
                        au=st1;ad=j;
                        al=tr[1].L,ar=tr[1].R;
                    }
                }
            }
        }
        while(i<n&&A[i+1].a==st1)i++;
    }
}
struct ls
{
    ll x,id;
}AA[1600];
bool cmp1(ls n1,ls n2){return n1.x<n2.x;}
int main()
{
    da*=1000000000;
    scanf("%lld",&n);ll z;
    for(ll i=1;i<=n;i++)scanf("%lld%lld%lld",&A[i].a,&A[i].f,&A[i].s);
    for(ll i=1;i<=n;i++)AA[i].x=A[i].a,AA[i].id=i;
    sort(AA+1,AA+n+1,cmp1);z=0;
    for(ll i=1;i<=n;i++)
    {
        if(i==1||AA[i].x!=AA[i-1].x){z++;ys1[z]=AA[i].x;}
        A[AA[i].id].a=z;
    }z1=z;
    for(ll i=1;i<=n;i++)AA[i].x=A[i].f,AA[i].id=i;
    sort(AA+1,AA+n+1,cmp1);z=0;
    for(ll i=1;i<=n;i++)
    {
        if(i==1||AA[i].x!=AA[i-1].x){z++;ys2[z]=AA[i].x;}
        A[AA[i].id].f=z;
    }z2=z;
    sort(A+1,A+n+1,cmp);
    ans=-da;solve();
    printf("%lld %lld\n%lld %lld\n",ys1[au],ys1[ad],ys2[al],ys2[ar]);
    return 0;
}

?

#include <iostream>
#include <string>

int main() {
    int n, gCount = 2;
    std::cin >> n;
    std::string guest;

    while (n--) {
        std::cin >> guest;
        gCount++;
        if (guest.find("+one", 0) != std::string::npos) gCount++;
    }
    if (gCount == 13) gCount++;
    std::cout << gCount * 100 << std::endl;

    return 0;
}

#include <iostream>
#include <vector>
#include <iterator>

int main() {
    int N;
    std::cin >> N;
    char v;
    int h, c = 0;
    std::vector<int> result;
    while (N--) {
        std::cin >> v >> h;
        v = v - 'a' + 1;
        if (h > 1 && v > 2) c += 1;
        if (h > 1 && v < 7) c += 1;
        if (h > 2 && v > 1) c += 1;
        if (h > 2 && v < 8) c += 1;

        if (h < 8 && v > 2) c += 1;
        if (h < 8 && v < 7) c += 1;
        if (h < 7 && v > 1) c += 1;
        if (h < 7 && v < 8) c += 1;

        result.push_back(c);
        c = 0;
    }
    std::copy(result.begin(), result.end(), std::ostream_iterator<int>(std::cout, "\n"));

    return 0;
}

I get AC with using algorithm with O(n) complexity - just use sliding window

Edited by author 28.07.2021 23:07

help me sol this problem
my email : npsdn1906@gmail.com

1) Try writing an algorithm for finding the minimum difference in a list first (list.sort() method may come in handy)

2) Think of how the usage of sets and lists may assist you in determining whether we have doublets in your list

3) Use counters for finding the number of iterations in your cycle

Enjoy :)

Edited by author 25.07.2021 21:52

Edited by author 25.07.2021 21:52

I think it's easy problem and must be rated 30

check your last string after FOR. I missed this but now Accepted.

You can use pbds_tree to solve this problem. PBDS Tree function named "order_by_key" will solve the problem like: find number of elements, which lower than x

what is #12 test?

#include<iostream>
#include<iomanip>
#include<math.h>
using namespace std;
double a[10001];
int main()
{
    int l=0,i,n,k;
    double x,m;
    cin>>n>>m;
    for(i=0;i<m;i++)
    {
    cin>>k;
    a[k]++;
    }
    for(i=1;i<=n;i++)
    {
    x=a[i]/m;
    x*=100;
    if(ceil(x)==floor(x))cout<<x<<".00%"<<endl;
    else
    cout<<setprecision(4)<<x<<'%'<<endl;
    }
    return 0;
}

Edited by author 09.04.2008 22:50

7 4
3 7 5 3

Check if the order RTLB is respected and pay attention to the fact that the matrix in the problem is inversed in comparison to the one that you build i.e. the bottom is curr_row - 1 and top is curr_row + 1.

whats problem on test 3??
что за проблема на 3 тесте??

Notice that "Filling value” is THREE decimal places.

Why always getting mle on #2