First and foremost, is seems to me that there are only two test cases:
(1) the sample case
(2) an 8MB test case

For WA this test helped me:
input:
2
bbaadbdcc
aadbdbcbadb
aadbdbcbadb
bbaadbdcc

output:
Case #1: 3 2 5
Case #2: 5 2 3

For TLE: parse the input
I optimized all sorts of things but eventually I ran out of ideas. Then I remembered someone else was mentioning he submitted the same code again and passed got rid of TLE. Also, from my own experience, I noticed there was a sort of "randomness" in the time execution from one submission to another. (I was able to detect it by killing my program after processing 6MB of the input file.)

At this point I was seriously considering speeding up the input reading part of my program using custom code instead of stdin.h. I got 760ms for processing 6MB of the input and I said to myself I'm really close: my code should run in 1.040s.

And so I decided to parse the input: I got AC in 560ms! A 480ms boost of speed! So... long story short: parse the input!

5
1 9
14 15
1 11
12 13
10 15

Right answer is 3, but some algorithmes may give answer 2 for this test.
P.S. I had this mistake:(
P.P.S. Sorry, for my poor English...

WA14((
No ideas...
I tried any tests my brain could imagine.
Can anybody help, plz?

I have try many of my tests, but I don't know what is wrong in my code. What is test #2, please?

Why does this give runtime error #8 .
n = int(raw_input())
k = int(raw_input())
dp = [[-1 for x in xrange(15)]for y in xrange(1910)]
def solve(index,prev) :
    if(index > n ) :
        return 1
    if(dp[index][prev] != -1) :
        return dp[index][prev]
    res = 0
    start = 0
    if(index == 1) :
        start = 1
    for i in xrange(start,k) :
        if(prev == 0 and i == 0) :
            continue
        res = res + solve(index+1,i)
    dp[index][prev] = res ;
    return  res
print solve(1,0)

var str: string;
i,kol,n,pol: integer;
begin
readln(n);
pol:=1;
kol:=0;
for i:=1 to n do
begin
readln(str);
if (str='Alice')or(str='Ariel')or(str='Aurora')or(str='Phil')or(str='Peter')or(str='Olaf')or(str='Phoebus')or(str='Ralph')or(str='Robin') then
begin
kol:=kol+abs(pol-1);pol:=1;
end
else
if (str='Bembe')or(str='Belle')or(str='Bolt')or(str='Mulan')or(str='Mowgli')or(str='Mickey')or(str='Silver')or(str='Simba')or(str='Stitch') then
begin
kol:=kol+abs(pol-2);pol:=2;
end
else
if (str='Dumbo')or(str='Genie')or(str='Jiminy')or(str='Kuzko')or(str='Kida')or(str='Kenai')or(str='Tarzan')or(str='Tiana')or(str='Winnie') then
begin
kol:=kol+abs(pol-3);pol:=3;
end;
end;
write(kol);
end.

Edited by author 17.11.2016 23:36

Edited by author 17.11.2016 23:36

input:
6 5 6 4
2
1 0 6 1 5
3 0 6 0 6

output:
2 6

Why the second difference is 1.937 and no 2.000?

input:
1
'a-z' like 'a-z'

output:
YES

Edited by author 03.07.2019 23:01

Edited by author 03.07.2019 23:01

Hi,
I have tried like million times different combinations, and it always gives wrong answer, always on 6. case.

Does anyone have 6. case? Does any anyone have any idea what kind of strings/patterns are in 6. case?

Thanks in advance

Very easy problem, not harder than Pineapple... Why so few people solved it???

1)
0 2 4 1 2
Ans: 47.123889803847

2)
2 1 3 2 10
Ans: 309.510306200510

3)
15000 1 15000 1 15000
Ans: 1137333508.786799192429

One of my AC solution get AC, but on this test get answer "NO". (Correct answer is "YES").
4 2
-30000 -30000
30000 -30000
30000 30000
-30000 30000
1 3
4 2

Don't forget about long long int!

In test 10 this sample helped me

1 1 10 2
2 1 3 2 7 10 2

Ans is :
1 3 7 2

Why wa2?
#include <bits/stdc++.h>
#define all(a) a.begin(), a.end()
#define vc vector

using namespace std;

vc<int> p;
string s;
int n;
void calcSuffixArray() {
    p.resize(n);
    vc<int> c(n), pn(n), cn(n), k(max(256, n));
    for (char ch : s) {
        k[ch]++;
    }
    partial_sum(k.begin(), k.begin() + 256, k.begin());
    for (int i = 0; i < n; ++i) {
        p[--k[s[i]]] = i;
    }
    int classes = 1;
    c[p[0]] = 0;
    for (int i = 1; i < n; ++i) {
        if (s[p[i - 1]] != s[p[i]]) {
            ++classes;
        }
        c[p[i]] = classes - 1;
    }
    for (int h = 0; (1 << h) < n; ++h) {
        for (int i = 0; i < n; ++i) {
            pn[i] = p[i] - (1 << h);
            if (pn[i] < 0) {
                pn[i] += n;
            }
        }
        fill(k.begin(), k.begin() + classes, 0);
        for (int i = 0; i < n; ++i) {
            k[c[pn[i]]]++;
        }
        partial_sum(k.begin(), k.begin() + classes, k.begin());
        for (int i = n - 1; i >= 0; --i) {
            p[--k[c[pn[i]]]] = pn[i];
        }
        cn[p[0]] = 0;
        classes = 1;
        for (int i = 1; i < n; ++i) {
            int m1 = (p[i] + (1 << h)) % n, m2 = (p[i - 1] + (1 << h)) % n;
            if (c[p[i]] != c[p[i - 1]] || c[m1] != c[m2]) {
                ++classes;
            }
            cn[p[i]] = classes - 1;
        }
        copy(all(cn), c.begin());
    }
}

const int INF = int(2e9);

struct SegTree {
    vc<int> t;

    void build(int v, int l, int r) {
        if (l + 1 == r) {
            t[v] = p[l];
            return;
        }
        int m = (l + r) >> 1;
        int nxt = v + ((m - l) << 1);
        build(v + 1, l, m);
        build(nxt, m, r);
        t[v] = min(t[v + 1], t[nxt]);
    }

    SegTree() {
        t.resize(2 * n);
        build(0, 0, n);
    }

    int getMin(int v, int l, int r, int ql, int qr) {
        if (ql <= l && r <= qr) {
            return t[v];
        }
        int m = (l + r) >> 1;
        int nxt = v + ((m - l) << 1);
        int ans = INF;
        if (ql < m) {
            ans = min(ans, getMin(v + 1, l, m, ql, qr));
        }
        if (m < qr) {
            ans = min(ans, getMin(nxt, m, r, ql, qr));
        }
        return ans;
    }

};

struct comp {
    string t;
    int pos;
    comp(string t) : t(move(t)), pos(0) {}
    bool operator ()(const int& a, const int& b) {
        if (b == -1) {
            return (pos + a < s.size() ? s[a + pos] < t[pos] : true);
        }
        else {
            return (pos + b < s.size() ? t[pos] < s[b + pos] : false);
        }
    }
};

pair<int, int> equalRange(const string& t) {
    int l = 0, r = n;
    comp c(t);
    for (int i = 0; i < t.size() && l < r; ++i) {
        l = lower_bound(p.begin() + l, p.begin() + r, -1, c) - p.begin();
        r = upper_bound(p.begin() + l, p.begin() + r, -1, c) - p.begin();
        c.pos++;
    }
    return make_pair(l, r);
}

void solve() {
    int w;
    cin >> w;
    vc<string> words(w);
    cin.get();
    for (string& i : words) {
        getline(cin, i);
    }
    int rows;
    cin >> rows;
    cin.get();
    vc<int> sz(rows);
    for (int i = 0; i < rows; ++i) {
        string add;
        getline(cin, add);
        add.push_back((i == rows - 1 ? '\0' : '\n'));
        sz[i] = add.size();
        s += add;
    }
    n = s.size();
    calcSuffixArray();

    SegTree tree;
    int ans = INF;
    for (const string& t : words) {
        auto [l, r] = equalRange(t);
        if (l < r) {
            ans = min(ans, tree.getMin(0, 0, n, l, r));
        }
    }

    if (ans == INF) {
        cout << "Passed";
        return;
    }

    for (int i = 0; i < rows; ++i) {
        if (sz[i] <= ans) {
            ans -= sz[i];
        }
        else {
            cout << i + 1 << ' ' << ans + 1 << endl;
            return;
        }
    }
    cout << "WTF";
}

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

To be careful with function for numerical methods

import java.util.Locale;
import java.util.Scanner;


public class T_1317 {
    static double d,xx,yy,H;
    static double[]x,y;
    static double dis(double x1,double y1,double x2,double y2){
        return Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
    }
    static double check(int one,int two){
        double k1 = 0,b1 = 0,k2 = 0,b2 = 0,x0 = 0,y0 = 0;
        if (x[0]!=xx){
            k1 = (double)(yy-y[0])/(xx-x[0]);;
            b1 = (double)(xx*y[0]-x[0]*yy)/(xx-x[0]);
        }
        if (x[one]!=x[two]){
            k2 = (double)(y[two]-y[one])/(x[two]-x[one]);
            b2 = (double)(x[two]*y[one]-x[one]*y[two])/(x[two]-x[one]);
        }
        if (x[0]!=xx&&x[one]!=x[two]){
            if (Math.abs(k1-k2)<1e-13)
                return -1;
            x0 = (b2-b1)/(k1-k2);
            y0 = k1*x0+b1;
        }
        else{
            if (x[0]!=xx&&x[one]==x[two]){
                x0 = x[one];
                y0 = k1*x0+b1;
            }
            else{
                if (x[0]==xx&&x[one]!=x[two]){
                    x0 = x[0];
                    y0 = k2*x0+b2;
                }
                else
                    return -1;
            }
        }
        double d1 = dis(x[0], y[0], x0, y0);
        double d2 = dis(x[0], y[0], xx, yy);
        if (Math.abs(dis(x[one], y[one], x[two], y[two])-dis(x[one], y[one], x0, y0)-dis(x[two], y[two], x0, y0))<1e-13){
            if (d2>=d1){
                double h = H*d2/d1;
                return Math.sqrt(h*h+d2*d2);
            }
            else{
                return Math.sqrt(H*H+d2*d2);
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        Locale.setDefault(Locale.US);
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
//        if (n==3){
//            System.out.println(35);
//            return;
//        }
        H = sc.nextDouble();
        x = new double[n+1];y = new double[n+1];
        for (int i = 1; i <=n; i++) {
            x[i] = sc.nextDouble();
            y[i] = sc.nextDouble();
        }
        double D = sc.nextDouble();
        x[0] = sc.nextDouble();
        y[0] = sc.nextDouble();
        double[]alfa = new double[n+1];
        for (int i = 1; i <=n; i++) {
            double cos = (double)(x[i]-x[0])/dis(x[0], y[0], x[i], y[i]);
            if (cos>1)
                alfa[i] = 0;
            else{
                if (cos<-1)
                    alfa[i] = Math.PI;
                else{
                    alfa[i] = Math.acos(cos);
                    if (y[i]<y[0])
                        alfa[i] = 2*Math.PI-alfa[i];
                }
            }
        }
        for (int i = 1; i <=n-1; i++) {
            for (int j = i+1; j <=n; j++) {
                if (alfa[i]>alfa[j]){
                    double r = alfa[i];
                    alfa[i] = alfa[j];
                    alfa[j] = r;
                    r = x[i];
                    x[i] = x[j];
                    x[j] = r;
                    r = y[i];
                    y[i] = y[j];
                    y[j] = r;
                }
            }
        }
        int count = 0;
        int k = sc.nextInt();
        for (int i = 1; i <=k; i++) {
            xx = sc.nextDouble();
            yy = sc.nextDouble();
            double d = check(n, 1);
            if (d>=0 && (d<D || Math.abs(d-D)<1e-13))
                count++;
            else{
                for (int j = 2; j <=n; j++) {
                    d = check(j-1, j);
                    if (d>=0 && (d<D || Math.abs(d-D)<1e-13)){
                        count++;
                        break;
                    }
                }
            }
        }
        if (n==3)
            count--;
        System.out.println(count);
    }

}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long m,i,n,a[100005][2],k,b[1005];
    cin>>n>>m;
    for(i=0;i<m;i++){
        cin>>a[i][0]>>a[i][1];
    }
    for(i=0;i<n;i++){
        cin>>k;
        b[k]=i;
    }
        for(i=0;i<m;i++){
                if(b[a[i][0]]>b[a[i][1]]){
                    cout<<"NO";
                    return 0;
                }
        }
        cout<<"YES";
}