I Count Two Three

最新推荐文章于 2020-02-20 23:22:41 发布

hang__19

最新推荐文章于 2020-02-20 23:22:41 发布

阅读量795

点赞数

分类专栏：比赛

比赛专栏收录该内容

24 篇文章

订阅专栏

Problem Description

I will show you the most popular board game in the Shanghai Ingress Resistance Team.
It all started several months ago.
We found out the home address of the enlightened agent Icount2three and decided to draw him out.
Millions of missiles were detonated, but some of them failed.

After the event, we analysed the laws of failed attacks.
It's interesting that the

i -th attacks failed if and only if

i can be rewritten as the form of

2a3b5c7d which

a,b,c,d are non-negative integers.

At recent dinner parties, we call the integers with the form

2a3b5c7d "I Count Two Three Numbers".
A related board game with a given positive integer

n from one agent, asks all participants the smallest "I Count Two Three Number" no smaller than

n .

Input

The first line of input contains an integer

t (1≤t≤500000) , the number of test cases.

t test cases follow. Each test case provides one integer

n (1≤n≤109) .

Output

For each test case, output one line with only one integer corresponding to the shortest "I Count Two Three Number" no smaller than

n .

Sample Input

Sample Output

  
  
   
   1
12
14
125
1250
12348
123480
1234800
12348000
123480000
  
  




#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=10000000000;
int a[112345];
int main()
{
    int i,j,k,e;
    long long x1,x2,x3,y;
    int top=0;
    for(i=0;i<30;i++)
    {
        x1=pow(2.0,i);
        for(j=0;j<19;j++)
        {
            x2=pow(3.0,j)*x1;
            if(x2>maxn) continue;//x2的值大于maxn时结束本次';
            for(e=0;e<13;e++)
            {
                x3=pow(5.0,e)*x2;
                if(x3>maxn) continue;
                for(k=0;k<11;k++)
                {
                    y=x3*pow(7.0,k);
                    if(y<=maxn)
                    a[top++]=y;//储存;
                }
            }
        }
    }
    sort(a,a+top);
    int t,p;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        p=upper_bound(a,a+top,n)-a;//找第一个比n大的数;
       if(a[p-1]==n) p--;//找第一个不小于n的数字;
        cout<<a[p]<<endl;
    }
}





*#include <iostream>
#include<math.h>
#include<algorithm>
using namespace std;
long long s[8100000],ss;
int i;


void f()
{
    int a,b,c,d;
    for(a=0;a<30;a++)
    {
        ss=pow(2,a);
        if(ss/100000000000<1)
            for(b=0;b<19;b++)
        {
            ss=pow(2,a)*pow(3,b);
            if(ss/100000000000<1)
                for(c=0;c<16;c++)
            {
                ss=pow(2,a)*pow(3,b)*pow(5,c);
                if(ss/10000000000<1)
                    for(d=0;d<12;d++)
                {
                    ss=pow(2,a)*pow(3,b)*pow(5,c)*pow(7,d);
                    if(ss/10000000000<1)
                        s[i++]=ss;
                    else
                        break;
                }
                else
                    break;
            }
            else
                break;
        }
        else
            break;
    }
}
int main()
{
    int n;
    long long x;
    i=0;
    f();
    sort(s,s+i);
    cin>>n;
    while(n--)
    {
        cin>>x;
        int p=lower_bound(s,s+i,x)-s;
        cout<<s[p]<<endl;
    }
    return 0;
}
#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <climits>
#include <ctime>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread(ch) freopen(ch,"r",stdin)
#define fwrite(ch) freopen(ch,"w",stdout)


using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const double eps = 1e-8;
const int maxn = 6000;
const LL bnd = 1123456789ll;


LL lis[maxn];
int len;


const int fac[4] = {2,3,5,7};


void dfs(int pos,LL v){
    if(pos == 4){
        lis[len++] = v;
        return;
    }
    do{
        dfs(pos+1,v);
        v *= fac[pos];
    }while(v <= bnd);
}


int main()
{
    //fread();
    //fwrite();
    len = 0;
    dfs(0,1);
    sort(lis,lis+len);
    int T;
    LL n;
    scanf("%d",&T);
    while(T-- && ~scanf("%I64d",&n)){
        printf("%I64d\n",*lower_bound(lis,lis+len,n));
    }
    return 0;
}
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N=110000;
int a[N];
int main(){
    int t;
    int j=1,k=1,x=1,y=1;
    a[1]=1;
    for(int i=2;i<=5843;i++){
        a[i]=min(2*a[k],min(3*a[j],min(5*a[x],7*a[y])));
        if(a[i]==2*a[k])k++;
        if(a[i]==3*a[j])j++;
        if(a[i]==5*a[x])x++;
        if(a[i]==7*a[y])y++;
    }
  //  cout<<a[5840];
//    for(int i=1;i<=10;i++){
//        cout<<a[i]<<endl;
//    }
    scanf("%d",&t);
    while(t--){
        int d;
        scanf("%d",&d);
        //cout<<d;
        int tt=lower_bound(a+1,a+5843,d)-a;
        printf("%d\n",a[tt]);
    }
    return 0;
}
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<climits>


using namespace std;
#define LL long long
#define maxn 1100000
LL  arr[maxn];
const LL MAX=2147483647;
LL vis[4][105];
int main(){
    vis[0][0] = vis[1][0] = vis[2][0] = vis[3][0] = 1;
    for(int i = 1;i < 33;i++)
        vis[0][i] = vis[0][i - 1] * 2;
    for(int i = 1;i < 26;i++)
        vis[1][i] = vis[1][i - 1] * 3;
    for(int i = 1;i < 19;i++)
        vis[2][i] = vis[2][i - 1] * 5;
    for(int i = 1;i < 16;i++)
        vis[3][i] = vis[3][i - 1] * 7;


    int t=0;
    LL temp;
    for(int i=0;i<33;++i){
        temp = vis[0][i];
        if(temp > MAX)
                continue;
        for(int j=0;j<26;++j){
            temp = vis[0][i] * vis[1][j];
            if(temp > MAX)
                continue;
            for(int k=0;k<19;++k){
                temp = vis[0][i] * vis[1][j] * vis[2][k];
                if(temp>MAX)
                    continue;
                for(int s=0;s<16;++s){
                    temp = vis[0][i] * vis[1][j] * vis[2][k] * vis[3][s];
                    if(temp>MAX)
                        continue;
                    else
                        arr[t++]=temp;
                }
            }
        }
    }
    sort(arr,arr+t);
    LL n,T;
    scanf("%I64d",&T);
    while(T-- && scanf("%I64d",&n)){
        int pos=lower_bound(arr,arr+t,n)-arr;
        printf("%I64d\n",arr[pos]);
    }
    return 0;
}*/