1 17
a 100 8

            long min = n;
            for (int i = 0; i < k; i++)
                for (int j = i + 1; j < k; j++) {
                    long diff = (data[i] + data[j] - n);
                    if (diff < min)
                        min = diff;
                }
            Console.WriteLine(min < 0 ? 0 : min);

Just use __int64 instead of long long. That helped me.

the original answer of fskdkskylp is ansthisone but my accepted code give this answer aêsthisone.


my accepted code


#include<bits/stdc++.h>
using namespace std;
int main()
{
    char c[104];
    scanf("%s",c);
    int l=strlen(c);
    int a[l+5];
    int b[l+5];
    for(int i=0; i<l; i++)
    {
        a[i]=c[i]-'a';
    }
    b[0]=a[0];
    if(a[0]<5)
    {
        b[0]=b[0]+26;
    }
    for(int i=1; i<l; i++)
    {
        int j=1;
        while((b[i]=a[i]+j*26))
        {
            if(b[i]%26==a[i]&& b[i]>=b[i-1])
                break;
            j++;
        }
    }
    printf("%c",b[0]-5+'a');
    for(int i=1; i<l; i++)
    {
        int j=abs(b[i]-b[i-1]);
        printf("%c",j+'a');
    }
    printf("\n");
    return 0;
}

9 8
1 2
1 6
1 8
1 3
1 4
4 5
1 7
7 9
Answer:
3
7 9
1 2
4 5

If you've got WA #11 then your program deals bad with repeatative queries.

Try this test:
--------------------
2
acm 10000
acmicpc 10000
2
ac
ac
---------------------
Correct answer is:
---------------------
acm
acmicpc

acm
acmicpc
---------------------

Is it true that k=f(n), where f(n) is Euler function?

Edited by author 13.02.2009 22:03

Answers for big tests is very big. For example, for test "50 500" answer is "854559745684320697549060368131279814466643179689928095831053239604130293492672614469791533133321".
Don't use long long or int64! Use long numbers (long arithmetic)! (Длинная арифметика)

It's really very easy problem! Just shuffle!

#include<bits/stdc++.h>
using namespace std;
int arr[15000],newarr[15000];
void sieve()
{
    arr[0]=1;
    arr[1]=1;
    newarr[1]=2;
    for(int i=4; i<15000; i+=2)
        arr[i]=1;
    for(int i=3; i<15000; i+=2){
        if(arr[i]!=1){
            for(int j=i*2; j<15000; j+=i){
                arr[j]=1;
            }
        }
    }
}

int main()
{
    int n, t, k=1;
    sieve();
    //cin>>n;
    for(int i=1; i<15000; i++){
        if(arr[i]!=1){
            newarr[k]=i;
            k++;
        }
    }
    cin>>t;
    while(t--){
        cin>>n;
        cout<<newarr[n]<<endl;
    }

    return 0;
}

Calculate the signed area of the polygon...

//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("avx")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

#define re return
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sqrt(x) sqrt(abs(x))
#define mp make_pair
#define pi (3.14159265358979323846264338327950288419716939937510)
#define fo(i, n) for(int i = 0; i < n; ++i)
#define ro(i, n) for(int i = n - 1; i >= 0; --i)
#define unique(v) v.resize(unique(all(v)) - v.begin())

template <class T> T abs (T x) { re x > 0 ? x : -x; }
template <class T> T sqr (T x) { re x * x; }
template <class T> T gcd (T a, T b) { re a ? gcd (b % a, a) : b; }
template <class T> int sgn (T x) { re x > 0 ? 1 : (x < 0 ? -1 : 0); }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef double D;
typedef long double ld;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef unsigned long long ull;
typedef tree <pair<int, char>, null_type, less<pair<int, char>>, rb_tree_tag, tree_order_statistics_node_update> _tree;

vector< tuple <int, int, string> > v;
vs ans;

int calc() {
    string ans;
    cin >> ans;
    re (ans[0] - '0') * 6000 + (ans[1] - '0') * 600 + (ans[3] - '0') * 100 + (ans[4] - '0') * 10 + (ans[6] - '0');
}

int main() {
    int best = (int) 1e9, n, cur;
    cin >> n;
    string str;
    fo(i, n) {
        cin >> str;
        cur = calc();
        v.pb(make_tuple(cur + i * 300, cur, str));
    }
    sort(all(v));
    int finish, res;
    fo(i, v.size()) {
        tie(finish, res, str) = v[i];
        if (res < best) {
            best = res;
            ans.pb(str);
        }
    }
    sort(all(ans));
    cout << ans.size() << endl;
    for (auto &j : ans) cout << j << '\n';
    re 0;
}

Edited by author 22.06.2018 17:22

I wrote suffix automaton, and with it's help i found solution in 5 lines.

It's unbelieveable, how couldn't I solve it during the contest.
Just brute-force for finding answers for smalls n...

I can't belive it!
    for(i=1;i<=n;i++){
        for(j=1;j<=n;j++){
            if(can[i][j]) for(k=1;k<=n;k++) if(can[j][k]) can[i][k]=1;
        }
    }

//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("avx")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

#define re return
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sqrt(x) sqrt(abs(x))
#define mp make_pair
#define pi (3.14159265358979323846264338327950288419716939937510)
#define fo(i, n) for(int i = 0; i < n; ++i)
#define ro(i, n) for(int i = n - 1; i >= 0; --i)
#define unique(v) v.resize(unique(all(v)) - v.begin())

template <class T> T abs (T x) { re x > 0 ? x : -x; }
template <class T> T sqr (T x) { re x * x; }
template <class T> T gcd (T a, T b) { re a ? gcd (b % a, a) : b; }
template <class T> int sgn (T x) { re x > 0 ? 1 : (x < 0 ? -1 : 0); }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef double D;
typedef long double ld;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef unsigned long long ull;
typedef tree <pair<int, char>, null_type, less<pair<int, char>>, rb_tree_tag, tree_order_statistics_node_update> _tree;

void Z_function (string &str, vi &z) {
    int n = str.size();
    z.resize(n, 0);
    for (int i = 1, l = 0, r = 0; i < n; ++i) {
        if (i <= r) {
            z[i] = min(r - i + 1, z[i - l]);
        }
        while (i + z[i] < n && str[z[i]] == str[i + z[i]]) ++z[i];
        if (i + z[i] - 1 > r) l = i, r = z[i] + i - 1;
    }
}

string str, s;
vi z;

int main() {
    int n;
    cin >> n >> str >> s;
    str += '$'; str += s; str += s;
    Z_function(str, z);
    fo(i, (int)z.size()) {
        if (z[i] == n) {
            cout << i - n - 1 << endl;
            re 0;
        }
    }
    cout << -1 << endl;
    re 0;
}

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <conio.h>

int main()
{
    int N;
    int i;
    int emp = 0, mac = 0, lit = 0;

    scanf("%i",&N);
    char **str = (char**)malloc(N * sizeof(char));
    for(i = 0; i < N; i++)
        str[i] = (char*)malloc(20);

    fflush(stdin);
    for(i = 0; i < N; i++)
        gets(str[i]);

    for(i = 0; i < N; i++)
    {
        if(strcmp(str[i], "Emperor Penguin") == 0)
            emp++;
        if(strcmp(str[i], "Macaroni Penguin") == 0)
            mac++;
        if(strcmp(str[i], "Little Penguin") == 0)
            lit++;
    }
    if(emp > mac && emp > lit)
        printf("%s\n", "Emperor Penguin");
    if(mac > emp && mac > lit)
        printf("%s\n", "Macaroni Penguin");
    if(lit > mac && lit > emp)
        printf("%s\n", "Little Penguin");

    for(i = 0; i < N; i++)
        free(str[i]);
    free (str);
    return 0;
}

#include <iostream>

using namespace std;

int main()
{
    int n;
    int c = 0;
    cin >> n;
    if (n>0)
    {
        for (int i = 1; i <= n; i++)
            c += i;
        cout << c;
    }
    else if (n <= 0)
    {
        for (int i = n; i <= 1; i++)
            c += i;
        cout << c;
    }
    return 0;
}