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

1

22

92

987654304

3825876150

Source

dp的思路很容易看出來，關鍵是處理好細節，調試了很長時間的代碼。

#include

#include

#include

#include

#include

#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<

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;

}

原文：http://blog.csdn.net/crper/article/details/20155953

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