PE 112 Bouncy numbers （模拟）-蒲公英云

PE 112 Bouncy numbers （模拟）

亦凉 2022-07-13 06:00 240阅读 0赞

Bouncy numbers

Problem 112

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a “bouncy” number; for example, 155349.

Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.

Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.

Find the least number for which the proportion of bouncy numbers is exactly 99%.

题意：额…自己看…

题解：额……手动模拟一遍就可以了….

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
int increase(ll x)
{
    int p=x%10;
    x/=10;
    while(x>0)
    {
        int c=x%10;
        if(c>p)return 0;
        p=c;
        x/=10;
     } 
     return 1;
}
int decrease(ll x)
{
    int p=x%10;
    x/=10;
    while(x>0)
    {
        int c=x%10;
        if(c<p)return 0;
        p=c;
        x/=10;
     } 
     return 1;
}
int gao(ll x)
{
    if(increase(x) || decrease(x))return 0;
    else return 1;
}
int main()
{
    ld s=0;
    for(int i=100;i<=10000000;i++)
    {
        if(gao(i)) s++;
        if(gao(i) && (s/i)*100.0==99)
        {
            cout<<i<<endl;
        }
    }
    return 0;
}