How many 0's? - POJ 3286 数位dp

最新推荐文章于 2020-09-01 10:43:24 发布

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本文介绍了一种计算在给定范围内所有自然数中零字符出现次数的算法。通过递归和动态规划的方法，该算法有效地解决了问题，并给出了完整的C++实现代码。

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How many 0's?

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2967		Accepted: 1582

Description

A Benedict monk No.16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0's will he write down?

Input

Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.

Output

For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.

Sample Input

10 11
100 200
0 500
1234567890 2345678901
0 4294967295
-1 -1

Sample Output

这道题和POJ2282一样，我就直接贴过来了。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
//dp[i][j]表示i位数中j的个数，dp0[i]表示i位数中合法的0的个数
ll dp[15][10],dp0[15],pow[15],n;
int num[15],len;
ll dfs(int pos,int N,int f)//f表示是否前面全是0，只对求0的个数有影响
{ if(pos<=0)
   return 0;
  ll ans=0;
  int i,j,k,m=num[pos];
  if(f==0)
  { ans+=m*dp[pos-1][N];
    if(N<m)
     ans+=pow[pos-1];
    else if(N==m)
     ans+=n%pow[pos-1]+1;
    ans+=dfs(pos-1,N,0);
  }
  else
  { ans+=(m-1)*dp[pos-1][N];
    ans+=dfs(pos-1,N,0);
    ans+=dp0[pos-1];
  }
  return ans;
}
ll solve(ll p,int N)
{ if(p<0)
   return -1;
  if(p==0)
   return 0;
  int i,j,k;
  len=0;n=p;
  while(p)
  { num[++len]=p%10;
    p/=10;
  }
  if(N==0)
   return dfs(len,N,1);
  else
   return dfs(len,N,0);
}
int main()
{ int i,j,k;
  ll l,r;
  pow[0]=1;
  for(i=1;i<=14;i++)
   pow[i]=pow[i-1]*10;
  for(i=1;i<=14;i++)
   for(j=0;j<=9;j++)
    dp[i][j]=dp[i-1][j]*10+pow[i-1];
  for(i=1;i<=14;i++)
   dp0[i]=9*dp[i-1][0]+dp0[i-1];
  while(~scanf("%I64d%I64d",&l,&r) && l!=-1)
  { if(l>r)
     swap(l,r);
    printf("%I64d\n",solve(r,0)-solve(l-1,0));
  }
}