CodeForces 374 B. Inna and Nine

最新推荐文章于 2019-08-03 11:19:00 发布

最新推荐文章于 2019-08-03 11:19:00 发布 · 132 阅读

Inna想要从一个包含数字1到9的数中最大化数字九的数量。通过选择相邻且和为9的两个数字并替换为单个9，她能实现这一目标。此挑战包括输入一个数a，并输出通过上述操作后能获得的最大数字九数量的唯一可能数。此外，解释了Inna如何将14545181变换为99919，以及解决此问题的策略。

Inna loves digit9very much. That's why she asked Dima to write a small number consisting of nines. But Dima must have misunderstood her and he wrote a very large numbera, consisting of digits from1to9.

Inna wants to slightly alter the number Dima wrote so that in the end the number contained as many digits nine as possible. In one move, Inna can choose two adjacent digits in a number which sum equals9and replace them by a single digit9.

For instance, Inna can alter number14545181like this:14545181 → 1945181 → 194519 → 19919. Also, she can use this method to transform number14545181into number19991. Inna will not transform it into149591as she can get numbers19919and19991which contain more digits nine.

Dima is a programmer so he wants to find out how many distinct numbers containing as many digits nine as possible Inna can get from the written number. Help him with this challenging task.

Input

The first line of the input contains integera(1 ≤ a ≤ 10100000). Numberadoesn't have any zeroes.

Output

In a single line print a single integer — the answer to the problem. It is guaranteed that the answer to the problem doesn't exceed263 - 1.

Please, do not use the%lldspecifier to read or write 64-bit integers in С++. It is preferred to use thecin,coutstreams or the%I64dspecifier.

Sample test(s)

Note

Notes to the samples

In the first sample Inna can get the following numbers:369727 → 99727 → 9997,369727 → 99727 → 9979.

In the second sample, Inna can act like this:123456789987654321 → 12396789987654321 → 1239678998769321.

xyxyxy....xy 时有1种变换方法

xyxyxy....xyx 时可以空出任意一个x

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>

using namespace std;

string a;

int main()
{
    unsigned long long int cnt=1;
    cin>>a;
    unsigned long long int len=a.size();
    unsigned long long int last=1;
    for(int i=0;i<len;i++)
    {
        if(a[i]=='9') continue;
        if(a[i]+a[i+1]==9+2*'0') last++;
        else if(last)
        {
            if(last%2) cnt*=(last+1)/2;
            last=1;
        }
    }
    cout<<cnt<<endl;
    return 0;
}