【HDU5878】I Count Two Three(打表+二分）

最新推荐文章于 2021-02-06 10:13:32 发布

原创最新推荐文章于 2021-02-06 10:13:32 发布 · 270 阅读

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本文介绍了一种算法，用于找到大于给定数值n的最小数，该数必须能表示为2^a * 3^b * 5^c * 7^d的形式。通过预先计算所有可能组合并排序来实现快速查询。

记录一个菜逼的成长。。

题目大意：
给你一个数n，求大于n的最小的满足2^a3^b5^c7^d的数。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <deque>
#include <cctype>
#include <bitset>
#include <cmath>
using namespace std;
#define ALL(v) (v).begin(),(v).end()
#define cl(a) memset(a,0,sizeof(a))
#define bp __builtin_popcount
#define pb push_back
#define fin freopen("D://in.txt","r",stdin)
#define fout freopen("D://out.txt","w",stdout)
#define lson t<<1,l,mid
#define rson t<<1|1,mid+1,r
#define seglen (node[t].r-node[t].l+1)
#define pi 3.1415926
#define e  2.718281828459
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef vector<PII> VPII;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
template <typename T>
inline void read(T &x){
    T ans=0;
    char last=' ',ch=getchar();
    while(ch<'0' || ch>'9')last=ch,ch=getchar();
    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
    if(last=='-')ans=-ans;
    x = ans;
}
inline bool DBread(double &num)
{
    char in;double Dec=0.1;
    bool IsN=false,IsD=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))
        in=getchar();
    if(in=='-'){IsN=true;num=0;}
    else if(in=='.'){IsD=true;num=0;}
    else num=in-'0';
    if(!IsD){
        while(in=getchar(),in>='0'&&in<='9'){
            num*=10;num+=in-'0';}
    }
    if(in!='.'){
        if(IsN) num=-num;
            return true;
    }else{
        while(in=getchar(),in>='0'&&in<='9'){
                num+=Dec*(in-'0');Dec*=0.1;
        }
    }
    if(IsN) num=-num;
    return true;
}
template <typename T>
inline void write(T a) {
    if(a < 0) { putchar('-'); a = -a; }
    if(a >= 10) write(a / 10);
    putchar(a % 10 + '0');
}
/******************head***********************/

LL mod_mul(LL a,LL b)
{
    LL res = 0;
    while(b){
        if(b & 1){
            res = (res + a);
        }
        b >>= 1;
        a = (a + a);
    }
    return res;
}
LL Power(LL a,LL n)
{
    LL ans = 1;
    while(n){
        if(n&1)ans *= a;;
        n /= 2;
        a *= a;
    }
    return ans;
}
const int maxn = 1000000 + 10;
LL num[maxn];
int main()
{
    LL a,b,c,d,tmp;
    int ind = 0;
    //要保证单独的数的幂都要超过10^9
    for( int i = 0; i < 31; i++ ){
    for( int j = 0; j < 20; j++ ){
    for( int k = 0; k < 16; k++ ){
    for( int o = 0; o < 13; o++ ){
        a = Power(2,i),b = Power(3,j),c = Power(5,k),d = Power(7,o);
        tmp = mod_mul(a,b);
        tmp = mod_mul(tmp,c);
        tmp = mod_mul(tmp,d);
        num[ind++] = tmp;
    }
    }
    }
    }
    sort(num,num+ind);
    int T;
    scanf("%d",&T);
    while(T--){
        LL n;
        scanf("%lld",&n);
        int pos = lower_bound(num,num+ind,n) - num;
        printf("%lld\n",num[pos]);
    }
    return 0;
}