codeforces1202D Print a 1337-string ... thinking

URL: http://codeforces.com/problemset/problem/1202/D

Meaning of the questions:

$ $ A n-input, the output contains only a $ 1,3,7 $ string, the string has a sequence $ $ $ 1337 $ n-length does not exceed $ 1e5 $.

answer:

First, we know that $ C_ {2} ^ {n} = \ frac {n (n-1)} {2} $ so if it satisfies $ n = \ frac {x (x-1)} {2} $ a, $ 133 $ 3337 ..... output (containing a $ X $ $ $ 3) to. This is obvious. Then, if it is not satisfied. Really I can not think when this race, behind wanted to clear up. First, because they no longer meet the $ n = \ frac {x (x-1)} {2} $, we have found from recent and less than $ $ $ n-n-$ a $ \ frac {x (x-1)} {2} $ calculated $ rem = n- \ frac {x (x-1)} {2} $, $ 13,377 and construction string ...... ..... 7733 $ 337 (containing a $ $ $ REM 7 $ and $ k $ a $ 3 $). The $ k $ how it should be calculated? Sequences and the front configuration comprising $ REM $ subsequences, then the back-containing $ C_ {k} ^ {2} $ in front and rear depicting a $ 3 $ are $ 2 \ cdot k $ a, and then combined with the foregoing $ 133 $ and final $ 7 $, $ 1 from $ a, so $ n = n- \ frac {x (x-1)} {2} + C_ {k} ^ {2} +2 \ cdot k + 1 $ a, because $ n, x $ known, calculated $ k $.

AC Code:

#include <bits/stdc++.h>
using namespace std;
long long num[100005];
void init()
{
    for(long long i=1;i<100005;++i)
        num[i]=i*(i-1)/2;
}
int main()
{
    init();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        int pos;
        for(pos=1;num[pos]<=n;++pos);
            pos-=1;
        if(n-num[pos])
        {
            cout<<"133";
            for(int i=0;i<n-num[pos];++i)
                cout<<'7';
            for(int i=0;i<pos-2;++i)
                cout<< '3';
            cout<<'7';
            cout<<endl;
        }
        else
        {
            cout<<'1';
            for(int i=0;i<pos;++i)
                cout<<'3';
            cout<<'7';
            cout<<endl;
        }
    }
    return 0;
}