POJ 3286 How many 0's?

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题目大意：

计算[m,n]之间所有数字有多少个零。

解题思路：

可以用[0,m）之间和[0,n]之间有多少个零然后作差。

规律是计算所有位置在到当前数时有多少个零。

下面是代码：

#include <set>#include <map>#include <queue>#include <math.h>#include <vector>#include <string>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#define eps 1e-8#define pi acos(-1.0)#define inf 107374182#define inf64 1152921504606846976#define lc l,m,tr<<1#define rc m + 1,r,tr<<1|1#define iabs(x)  ((x) > 0 ? (x) : -(x))#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (SIZE))#define clearall(A, X) memset(A, X, sizeof(A))#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))#define memcopyall(A, X) memcpy(A , X ,sizeof(X))#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )using namespace std;long long does(long long num){    if(num<0)return 0;    long long ans=1,div=10;    while(num/div)    {        if((num%div-num%(div/10))==0)ans+=((num/div)-1)*(div/10)+(num%(div/10))+1;        else ans+=(num/div)*(div/10);        div*=10;    }    return ans;}int main(){    long long n,m;    while(scanf("%lld%lld",&m,&n),n!=-1&&m!=-1)    {        printf("%lld\n",does(n)-does(m-1));    }    return 0;}

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