标签：ons map integer ack ima any not numbers temp
Jimmy writes down the decimal representations of all natural numbers between and including m and n, (m ≤ n). How many zeroes will he write down?
Input
Input starts with an integer T (≤ 11000), denoting the number of test cases.
Each case contains two unsigned 32-bit integers m and n, (m ≤ n).
Output
For each case, print the case number and the number of zeroes written down by Jimmy.
Sample Input
5
10 11
100 200
0 500
1234567890 2345678901
0 4294967295
Sample Output
Case 1: 1
Case 2: 22
Case 3: 92
Case 4: 987654304
Case 5: 3825876150
题解：数位DP入门；我们先处理处数字的每一位，然后对每一位考虑，同时记录是否
有前导零，以及前导零的个数；
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<queue> #include<stack> #include<set> #include<vector> #include<map> #include<string> using namespace std; typedef long long LL; LL dp[40][40],digit[40],n,m; int T; LL dfs(int pos,int pre,int cnt,bool Judge) { if(pos==0) return cnt; int sz=Judge? digit[pos]:9; if(!Judge && dp[pos][cnt]!=-1 && pre) return dp[pos][cnt]; LL ans=0; for(int i=0;i<=sz;i++) { if(i==0&&pre) ans+=dfs(pos-1,1,cnt+1,Judge&&i==sz); else ans+=dfs(pos-1,i!=0||pre,cnt,Judge&&i==sz); } if(!Judge&&pre) dp[pos][cnt]=ans; return ans; } LL work(LL num) { int temp=0; if(num==-1) return -1; while(num) { digit[++temp]=num%10; num/=10; } return dfs(temp,0,0,1); } int main() { ios::sync_with_stdio(false); cin.tie(0); cin>>T; for(int i=1;i<=T;i++) { memset(dp,-1,sizeof dp); cin>>m>>n; cout<<"Case "<<i<<": "<<work(n)-work(m-1)<<endl; } return 0; }
Light oj 1140 How Many Zeroes?
标签：ons map integer ack ima any not numbers temp
原文地址：https://www.cnblogs.com/songorz/p/9452832.html
