codeforces 1348D (思维+贪心)

 

 

题意:一天分为白天和黑夜,白天可以选择让任意份细胞分解,一份细胞可分解为两份,重量平分。晚上每一份细胞重量+1。给一个n,初始有一份细胞,重量为1。问最少多少次可使得细胞总重量为n。

思路:倘若每天都让所有的细胞分裂,则每天加的cha值都会×2,定义一个cun数组,用来存每天新加的重量(既是cha值),sum表示当前这天细胞总质量为多少。

  当sum大于n,则表示这天的重量超过了,则需要前面抽出某一天k,让这一天cha值为sum-n。因为cha值是递增的,所以只需要在尾部添加然后sort即可。

  cun数组用来存新加的cha值,所以cun数组的差分数组是cha值的增加值即答案。

 

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '\n'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
int cun[100];
int main()
{
    //freopen("input.txt", "r", stdin);
    int _;
    scanf("%d",&_);
    while(_--)
    {
        ll n;
        cin>>n;
        int tail = 0;
        int cha = 1;
        ll sum = 1;
        cun[++tail] = 1;

        while(1)
        {
            cha *= 2;
            if(sum + cha == n){
                cun[++tail] = cha;    
                break;
            }
            else if(sum + cha > n){
                cun[++tail] = n-sum;
                break;
            }
            cun[++tail] = cha;
            sum += cha;
        }
        /*rep(i,1,tail) cout<<cun[i]<<sp;
        cout<<endl;*/
        sort(cun+1,cun+1+tail);
        cout<<tail-1<<endl;
        rep(i,2,tail){
            cout<<cun[i]-cun[i-1]<<sp;
        }
        cout<<endl;
    }
}

 

posted @ 2020-06-30 20:17  阿斯水生产线  阅读(190)  评论(0编辑  收藏  举报