51Node 1010 只包含因子2 3 5的数 思维

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题目

https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1010

题意

思路

首先枚举出所有满足条件且不超过 $10^{18}$的数字，然后二分。

代码

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;

const int N = 100000 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double PI = acos(-1);

LL arr[N];

int init(LL limit)
{
    int index2 = 0, index3 = 0, index5 = 0;
    int len = 0;
    arr[0] = 1;
    while(arr[len] <= limit)
    {
        arr[++len] = min({arr[index2]*2, arr[index3]*3, arr[index5]*5});
        while(arr[index2]*2 <= arr[len]) ++index2;
        while(arr[index3]*3 <= arr[len]) ++index3;
        while(arr[index5]*5 <= arr[len]) ++index5;
    }
    return len;
}

int main()
{
    int len = init(1e18 + 10);

    int T;
    LL n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld", &n);
        printf("%lld\n", *lower_bound(arr+1, arr+len+1, n));
    }

    return 0;
}

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;

const int N = 100000 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double PI = acos(-1);

LL arr[N];

int init(LL limit)
{
    int len = 0;
    for(LL i = 1; i < limit; i *= 2)
    {
        for(LL j = 1; i*j < limit; j *= 3)
        {
            for(LL k = 1; i*j*k < limit; k *= 5)
                arr[len++] = i*j*k;
        }
    }
    sort(arr, arr + len);
    return len;
}

int main()
{
    int len = init(1e18 + 100);//注意，这里偷懒用了浮点数形式，但超过15位后精度损失很大，填1e18+10得出的是10^18，填1e18+100得出的18^18+128
    //cout << arr[len-1] << endl;

    int T;
    LL n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld", &n);
        printf("%lld\n", *lower_bound(arr+1, arr+len, n));
    }

    return 0;
}