test

Just follow what Dinomax(orz) did.

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As everyone knows, the higher the concentration is, the stronger the man will be.

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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<set>
#include<map>
using namespace std;

//faster_reader
inline int read ()
{
    int f = 1, a = 0;
    char c = getchar ();
    while (c < '0' || c > '9')
    {
        if (c == '-') f = -1;
        c = getchar ();
    }
    while (c >= '0' && c <= '9')
    {
        a = a * 10 + c - '0';
        c = getchar ();
    }
    return f * a;
}

//high_accuracy
const int maxn = 100;
struct bign{
    int d[maxn], len;

    void clean() { while(len > 1 && !d[len-1]) len--; }

    bign()          { memset(d, 0, sizeof(d)); len = 1; }
    bign(int num)   { *this = num; }
    bign(char* num) { *this = num; }
    bign operator = (const char* num){
        memset(d, 0, sizeof(d)); len = strlen(num);
        for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';
        clean();
        return *this;
    }
    bign operator = (int num){
        char s[20]; sprintf(s, "%d", num);
        *this = s;
        return *this;
    }

    bign operator + (const bign& b){
        bign c = *this; int i;
        for (i = 0; i < b.len; i++){
            c.d[i] += b.d[i];
            if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++;
        }
        while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;
        c.len = max(len, b.len);
        if (c.d[i] && c.len <= i) c.len = i+1;
        return c;
    }
    bign operator - (const bign& b){
        bign c = *this; int i;
        for (i = 0; i < b.len; i++){
            c.d[i] -= b.d[i];
            if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;
        }
        while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;
        c.clean();
        return c;
    }
    bign operator * (const bign& b)const{
        int i, j; bign c; c.len = len + b.len;
        for(j = 0; j < b.len; j++) for(i = 0; i < len; i++)
            c.d[i+j] += d[i] * b.d[j];
        for(i = 0; i < c.len-1; i++)
            c.d[i+1] += c.d[i]/10, c.d[i] %= 10;
        c.clean();
        return c;
    }
    bign operator / (const bign& b){
        int i, j;
        bign c = *this, a = 0;
        for (i = len - 1; i >= 0; i--)
        {
            a = a*10 + d[i];
            for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
            c.d[i] = j;
            a = a - b*j;
        }
        c.clean();
        return c;
    }
    bign operator % (const bign& b){
        int i, j;
        bign a = 0;
        for (i = len - 1; i >= 0; i--)
        {
            a = a*10 + d[i];
            for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
            a = a - b*j;
        }
        return a;
    }
    bign operator += (const bign& b){
        *this = *this + b;
        return *this;
    }

    bool operator <(const bign& b) const{
        if(len != b.len) return len < b.len;
        for(int i = len-1; i >= 0; i--)
            if(d[i] != b.d[i]) return d[i] < b.d[i];
        return false;
    }
    bool operator >(const bign& b) const{return b < *this;}
    bool operator<=(const bign& b) const{return !(b < *this);}
    bool operator>=(const bign& b) const{return !(*this < b);}
    bool operator!=(const bign& b) const{return b < *this || *this < b;}
    bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);}

    string str() const{
        char s[maxn]={};
        for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';
        return s;
    }
};
istream& operator >> (istream& in, bign& x)
{
    string s;
    in >> s;
    x = s.c_str();
    return in;
}
ostream& operator << (ostream& out, const bign& x)
{
    out << x.str();
    return out;
}

//graph
const int N = 1010, M = 100010;
int head[N], num = 0;
struct edge
{
    int nxt, to, val;
};
edge e[M];
void add_edge(int x, int y, int z)
{
    e[++ num].nxt = head[x];
    head[x] = num;
    e[num].to = y;
    e[num].val = z;
}

//quick_pow
const int mod = 1e9 + 7;
long long qpow(long long x, int y)
{
    long long ans = 1;
    while (y)
    {
        if (y & 1)
        {
            ans *= x;
            ans %= mod;
        }
        x *= x;
        x %= mod;
        y >>= 1;
    }
    return ans;
}

//gcd
inline int gcd(int a, int b)
{
    int r;
    while(b > 0)
    {
        r = a % b;
        a = b;
        b = r;
    }
    return a;
}

int main()
{
    int T = 1;
    //T = read();
    while (T --)
    {

    }
    return 0;
}

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OK

It's time for codeforces