Phoenix and Science [CF1348 D] (Mathematics, Laws)

D. Phoenix and Science

CF1348 D
title meaning:
cell division, the number on the first day is $1$ . There can be0 0every day after that$0$ ~$Whenbeforethe\; finecellsanumber\; of$ cells in$m\; is$ split into$\frac{m}{2}$. Then every cell will be +1 +1 that night$+1$ .
Q becomesnnThe minimum number of days for$n$ cells.

Idea:
Observe the sample and push it with your hand.
Find:For the total, $m⟹\frac{m}{2}$Does not change the final $sum$ will only increase the night addend$ans$。
The largest addend $ans$ growth rate is$2^x$ increases exponentially, and the number of days is the smallest at this time.
But note that for the daily addend$For\; ans$ , this is a non-strictly increasing sequence, and the output value is the addendans ansThe difference of$a$$n$$s$ (greater than or equal to 0).
When calculating, the addend ans ans of the last day$ans$ is$n-2^i+1\; It\; is$ calculated that it may be less than theans ans of theprevious day$ans$ , remember to deal with the final day's addend$ans\; is$ greater than or equal to the day before

LL ans[100];
int main(){
    
    
    int t;
    cin>>t;
    LL n,nw,pre,tt;
    while(t--){
    
    
        cin>>n;
        ans[1]=1;
        nw=1;tt=2;
        for(int i=1;i<=50;i++){
    
    
            ans[i+1]=tt;
            if(tt*2-1>=n){
    
    
                ans[i+1]=n-tt+1;
                nw=i;
                break;
            }
            tt*=2ll;
        }
        cout<<nw<<endl;
        if(ans[nw+1]<ans[nw]){
    
    
            tt=ans[nw+1]+ans[nw];
            ans[nw]=tt/2;
            ans[nw+1]=tt/2+tt%2;
        }
        for(int i=2;i<=nw+1;i++){
    
    
            cout<<ans[i]-ans[i-1]<<" ";
        }
        cout<<endl;
    }
}