poj 3286 How many 0's? 按位统计

//poj 3286 //sep9 #include<iostream> using namespace std; typedef __int64 ll; ll b[16]; ll f(ll n) { ll left,m,ans=0; for(int i=1;i<13;++i){ left=n/b[i]⑴; ans+=left*b[i⑴]; m=(n%b[i]-n%b[i⑴])/b[i⑴]; if(m>0) ans+=b[i⑴]; else if(m==0) ans+=n%b[i⑴]+1; if(b[i]>n) break; } return ans; } int main() { ll m,n; b[0]=1; for(int i=1;i<=12;++i) b[i]=b[i⑴]*10; while(scanf("%I64d%I64d",&m,&n)&&n!=⑴){ printf("%I64d ",f(n)-f(m⑴)); } return 0; }