URAL 2047 Maths (数学)

对于一个数来说,它的除数是确定的,那么它的前驱也是确定的,而起点只能是1或2,所以只要类似筛法先预处理出每个数的除数个数

,然后递推出每个数往前的延伸的链长,更新最大长度,记录对应数字。找到maxn以后,根据最后一个数找到前驱,并记录到ans数组中。

代码来自队友

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<string>
#include<iostream>
#include<set>
#include<map>
#define mod  1000000007
#define inf 1000000001
#define big 1570000
using namespace std;

int d[big+1];
int dp[big+1];
int ans[big+1];

void find()
{

    for (int i = 1; i <big; i++)
    {
        for (int j = i; j <= big; j += i)
            d[j]++;
    }
}

int main()
{
    find();
    dp[1] = 1;
    dp[2] = 1;
    int maxn = 0;
    int re;
    for (int i = 3; i <=big; i++)
    {
        dp[i] = dp[i - d[i]] + 1;
        if (dp[i] > maxn)
        {
            maxn = dp[i];
            re = i;
        }
    }
    for (int i = maxn; i >= 0; i--)
    {
        ans[i] = d[re];
        re -= d[re];
    }
    int n;
    scanf("%d", &n);
    if (n > maxn)
        printf("Impossible
");
    else
    {
        for (int i = 1; i < n; i++)
        {
            printf("%d ", ans[i]);
        }
        printf("%d
", ans[n]);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/jerryRey/p/4719737.html