POJ 3286 How many 0's?

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最新推荐文章于 2025-03-30 16:45:00 发布

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How many 0's?

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2844		Accepted: 1506

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

Source

Waterloo Local Contest, 2006.5.27

dp的思路很容易看出来，关键是处理好细节，调试了很长时间的代码。

#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdio>
#define N 30
using namespace std;
__int64 dp1[N],dp2[N];
int dig[N];
int Top;
__int64 ba[N];
int main()
{
    //freopen("data.txt","r",stdin);
    __int64 f();
    memset(dp1,0,sizeof(dp1));
    memset(dp2,0,sizeof(dp2));
    dp1[1] = dp2[1] = 1;
    ba[0] = 1;
    ba[1] = 10;
    for(int i=2;i<=11;i++)
    {
        dp1[i] = dp2[i-1]*9;
        dp2[i] = 10*dp2[i-1]+ba[i-1];
        ba[i] = ba[i-1] * 10;
    }
    __int64 n,m;
    while(scanf("%I64d %I64d",&n,&m)!=EOF)
    {
        if(n<0)
        {
            break;
        }
        Top = 0;
        n-=1;
        __int64 res1,res2;
        if(n==-1)
        {
            res1 = 0;
        }else
        {
            if(n==0)
            {
                dig[Top++] = 0;
            }
            while(n!=0)
            {
               dig[Top++] = n%10;
               n = n/10;
            }
            res1=f();
        }
        Top = 0;
        if(m==0)
        {
            dig[Top++] = 0;
        }
        while(m!=0)
        {
            dig[Top++] = m%10;
            m = m/10;
        }
        res2 = f();
        printf("%I64d\n",res2-res1);
    }
    return 0;
}
__int64 f()
{
    __int64 res = 0;
    for(int i=1;i<=Top-1;i++)
    {
        res+=dp1[i];
    }
    //cout<<res<<endl;
    int t;
    for(int i=Top-1;i>=0;i--)
    {
        if(i==Top-1&&Top>1)
        {
            t = 1;
        }else
        {
            t = 0;
        }
        for(int j=t;j<=dig[i]-1;j++)
        {
            res+=dp2[i];
            if(j==0)
            {
                res+=ba[i];
            }
        }
    }
    for(int i=Top-1;i>=0;i--)
    {
        if(dig[i]==0)
        {
            __int64 sum = 0;
            for(int j=i-1;j>=0;j--)
            {
                sum = sum*10+dig[j];
            }
            sum+=1;
            res+=sum;
        }
    }
    return res;
}