POJ 3286- How many 0's?(组合数学_区间计数)

How many 0's?

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3286

Appoint description: System Crawler  (2015-04-18)

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 11100 2000 5001234567890 23456789010 4294967295-1 -1

Sample Output

122929876543043825876150

题意：求区间内出现多少个零。和POJ 2282差不多。

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const double pi= acos(-1.0);#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1LL b[12]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000,10000000000,100000000000};LL get_res(LL n){    int i;    LL left,m;    LL cnt=0;    for(i=1;i<12;i++){            left=n/b[i]-1;        cnt+=left*b[i-1];        m=(n%b[i]-n%b[i-1])/b[i-1];        if(m>0)            cnt+=b[i-1];        else if(m==0)            cnt+=n%b[i-1]+1;        if(n<b[i])            break;    }    return cnt;}int main(){    LL n,m;    while(~scanf("%lld %lld",&n,&m)){        if(n==-1&&m==-1) break;        if(n>m) swap(n,m);        printf("%lld\n",get_res(m)-get_res(n-1));    }    return 0;}