标签：submit += show mst std second normal style com
| Time Limit: 2 second(s) | Memory Limit: 32 MB |
A palindromic number or numeral palindrome is a ‘symmetrical‘ number like 16461 that remains the same when its digits are reversed. In this problem you will be given two integers i j, you have to find the number of palindromic numbers between i and j (inclusive).
Input starts with an integer T (≤ 200), denoting the number of test cases.
Each case starts with a line containing two integers i j (0 ≤ i, j ≤ 1017).
For each case, print the case number and the total number of palindromic numbers between i and j (inclusive).
Sample Input |
Output for Sample Input |
|
4 1 10 100 1 1 1000 1 10000 |
Case 1: 9 Case 2: 18 Case 3: 108 Case 4: 198 |
//开个数组存一下回文串，还要处理前导零。sta表示到目前为止是否合法 #include<iostream> #include<cstdio> #include<cstring> using namespace std; typedef long long ll; int t,bit[20],num[20]; ll f[20][20][2],P,Q; ll dfs(int pos,int s,int sta,bool limit) { if(pos==0) return sta; if(!limit&&f[pos][s][sta]!=-1) return f[pos][s][sta]; int max_b=(limit?bit[pos]:9); ll ans=0; for(int i=0;i<=max_b;i++){ num[pos]=i; if(pos==s&&i==0) ans+=dfs(pos-1,s-1,sta,limit&&(i==max_b)); else if(sta&&pos<=s/2) ans+=dfs(pos-1,s,num[s-pos+1]==i,limit&&(i==max_b)); else ans+=dfs(pos-1,s,sta,limit&&(i==max_b)); } if(!limit) f[pos][s][sta]=ans; return ans; } ll solve(ll x) { if(x<0) return 0; int pos=0; while(x){ bit[++pos]=x%10; x/=10; } return dfs(pos,pos,1,1); } int main() { memset(f,-1,sizeof(f)); scanf("%d",&t); for(int cas=1;cas<=t;cas++){ scanf("%lld%lld",&P,&Q); if(P>Q) swap(P,Q); printf("Case %d: %lld\n",cas,solve(Q)-solve(P-1)); } return 0; }
标签：submit += show mst std second normal style com
原文地址：http://www.cnblogs.com/--ZHIYUAN/p/7674243.html
