Codeforces Round #431 (Div. 2) 848A From Y to Y(思维)

今早补题感觉思路不错,这题昨天连看都没看,真是菜到无法想象了.
一开始看到,有点懵逼,觉得很难,因为有一些,exactly,minimal,这样的词语,搞得好像这题状况很多一样.
但是仔细看会发现,每一种字符串的形式,他的”操作值”,是固定的,那么,这题就不那么难了.
又分析发现,假如是aaaa这种,他们要合在一起,操作值一定是1+2+3=6,
aaaaa呢,一定是6+4 = 10,也就是一个阶和的情况.
那么这题就容易了,目的变为要把一个n,分解成多个阶和.
使用二分瞎逼写一写就过了.

/*  xzppp  */#include <iostream>#include <vector>#include <cstdio>#include <string.h>#include <algorithm>#include <queue>#include <map>#include <string>#include <cmath>#include <stack>#include <iomanip>using namespace std;#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define MP make_pair#define PB push_backtypedef long long  LL;typedef unsigned long long ULL;typedef pair<int,int > pii;typedef pair<double,double > pdd;typedef pair<double,int > pdi;const int MAXN = 1500+17;const int MAXM = 20;const int MAXV = 2*1e3+17;const int INF = 0x7fffffff;const int MOD = 1e9+7;int sum[MAXN];int main(){    #ifndef ONLINE_JUDGE     FFF    #endif    sum[2] = 1;    for (int i = 3; i < MAXN; ++i)    {        sum[i] = sum[i-1] + i-1;    }    int n;    cin>>n;    bool can = false;    vector<int > ans;    int tie = 0;    if(n==0)    {        cout<<"a"<<endl;        return 0;    }    while(!can)    {        if(tie++>100) break;        int pos = (lower_bound(sum, sum+MAXN, n))-sum-1;        int x = n;        ans.PB(pos);        int temp = pos;        x -= sum[temp];        while(x>0)        {            int temp = (lower_bound(sum, sum+MAXN, x))-sum;            if(sum[temp]>x) temp--;            x -= sum[temp];            ans.push_back(temp);        }        if(x==0)        {            can = true;            break;        }        else        {            ans.clear();        }    }    for (int i = 0; i < ans.size(); ++i)    {        for (int j = 0; j < ans[i]; ++j)        {            printf("%c",i+'a');        }    }    return 0;}