UVA 1640（DFS）

题意：给你a,b两个数 问你a b区间中0 9出现的次数

其实就是求1-n中0-9出现的次数 ans[n]   答案就是ans[b]-ans[a-1]

怎么求的话看代码吧

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<string.h>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<map>
 #include<cmath>
 typedef long long ll;
 typedef unsigned long long LL;
 using namespace std;
 const double PI=acos(-1.0);
 const double eps=0.0000000001;
 const int N=+;
 const ll mod=1e9+;
 const int INF=0x3f3f3f3f;
 int a[N];
 int b[N];
 int ans[N];
 void DFS(int n,int m){
     int x=n/;
     int y=n%;
     for(int i=;i<=;i++)ans[i]=ans[i]+x*m;
     for(int i=;i<=y;i++)ans[i]=ans[i]+m;
     if(x==)return;
     DFS(x-,m*);
     while(x){
         int t=x%;
         ans[t]=ans[t]+m*(y+);
         x=x/;
     }

 }
 int main(){
     int n,m;
     while(scanf("%d%d",&n,&m)!=EOF){
         if(n==&&m==)break;
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         memset(ans,,sizeof(ans));
         int t1=;
         int t2=;
         if(n>m)swap(n,m);
         DFS(n-,);
         for(int i=;i<=;i++)a[i]=ans[i];
         memset(ans,,sizeof(ans));
         DFS(m,);
         for(int i=;i<=;i++)b[i]=ans[i];
         for(int i=;i<=;i++)cout<<b[i]-a[i]<<" ";
         cout<<b[]-a[]<<endl;

     }
 }