LightOJ 1205 Palindromic Numbers - c++编程基础

数位DP。。。。

Palindromic Numbers

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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Description

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

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).

Output

For each case, print the case number and the total number of palindromic numbers between i and j (inclusive).

Sample Input

1 10

100 1

1 1000

1 10000

Sample Output

Case 1: 9

Case 2: 18

Case 3: 108

Case 4: 198

Source

Problem Setter: Jane Alam Jan

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<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+PHByZSBjbGFzcz0="brush:java;">#include #include #include #include using namespace std; typedef long long int LL; int a[70]; LL dp[70][70]; LL dfs(int len,int l,int r,bool limit,bool ok) { if(l =i; else g=a[r]>i; ret+=dfs(len,l-1,r+1,limit&&i==mx,g); } if(!limit) dp[len][l]=ret; return ret; } LL gaoit(LL n) { if(n<0) return 0; if(n==0) return 1; int len=0; while(n){a[len++]=n%10;n/=10;} LL ret=1; for(int i=len;i>=1;i--) ret+=dfs(i,i-1,0,i==len,1); return ret; } int main() { int T_T,cas=1; cin>>T_T; memset(dp,-1,sizeof(dp)); while(T_T--) { LL x,y; cin>>x>>y; if(x>y) swap(x,y); printf("Case %d: %lld\n",cas++,gaoit(y)-gaoit(x-1)); } return 0; }