CodeForces 617B 2

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最新推荐文章于 2024-09-28 09:20:52 发布

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分类专栏： CF 文章标签： codeforces

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本文解析了一道关于如何将含有坚果的巧克力条进行有效分割的算法题目，介绍了解题思路和实现代码，通过统计每两个坚果间的空白位置数量来确定分割方式。

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B - 2

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 617B

Description

Bob loves everything sweet. His favorite chocolate bar consists of pieces, each piece may contain a nut. Bob wants to break the bar of chocolate into multiple pieces so that each part would contain exactly one nut and any break line goes between two adjacent pieces.

You are asked to calculate the number of ways he can do it. Two ways to break chocolate are considered distinct if one of them contains a break between some two adjacent pieces and the other one doesn't.

Please note, that if Bob doesn't make any breaks, all the bar will form one piece and it still has to have exactly one nut.

Input

The first line of the input contains integer n (1 ≤ n ≤ 100) — the number of pieces in the chocolate bar.

The second line contains n integers a_i (0 ≤ a_i ≤ 1), where 0 represents a piece without the nut and 1 stands for a piece with the nut.

Output

Print the number of ways to break the chocolate into multiple parts so that each part would contain exactly one nut.

Sample Input

Input

3
0 1 0

Output

Input

5
1 0 1 0 1

Output

Hint

In the first sample there is exactly one nut, so the number of ways equals 1 — Bob shouldn't make any breaks.

In the second sample you can break the bar in four ways:

10|10|1

1|010|1

10|1|01

1|01|01

题意：给你一个数列，需要你来把这个数列分割，要求每一部分有且仅有1个1。

思路：每两个1之间的空都可以切，找到这些间隔，然后每个间隔里面的空相加，最后间隔之间相乘就可以了。

注意：如果1后面没有1了，那么这整个1000.....就只有一种切法了！

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int MAXN=100+5;
const int inf=1000000000;
int a[MAXN];
int sum[MAXN];
int main()
{
    int n;
    int i;
    scanf("%d",&n);
    for(i=0;i<n;++i)scanf("%d",&a[i]);
    i=0;
    int cnt=0;
    while(i<n)
    {
        if(a[i])
        {
            while(!a[++i]&&i<n-1){sum[cnt]++;}//记录0的数量
            if(!a[i]&&i==n-1)sum[cnt]=0;//如果后面全为0了就只有一种切法
            cnt++;
        }
        else i++;
    }
    if(cnt>0)//不全为0
    {
        long long ans=1;
    for(i=0;i<cnt;++i)
    {
        ans*=sum[i]+1;
    }
    printf("%lld\n",ans);
    }
    else puts("0");
    return 0;
}