标签：i++ open play pac case out continue isp mic
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).
InputInput 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).
OutputFor each case, print the case number and the total number of palindromic numbers between i and j (inclusive).
Sample Input4
1 10
100 1
1 1000
1 10000
Sample OutputCase 1: 9
Case 2: 18
Case 3: 108
Case 4: 198
题意：求区间的回文串数量。
思路：从两头向中间靠，前缀是否小于原数用tag表示，后缀是否小于原数用ok表示，注意后缀尽管后面的比原位大，但是前面的小一点可以抵消其效果。
#include<bits/stdc++.h> #define ll long long #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; ll dp[20][20][2][2]; int d[20],cnt; ll get(int bg,int l,int r,int tag,bool ok) { if(r>l) return !tag||(tag&&ok); if(!tag&&dp[bg][l][tag][ok]) return dp[bg][l][tag][ok]; int lim=tag?d[l]:9; ll res=0; rep(i,0,lim){ if(bg==l&&i==0) continue; bool g=ok; if(ok) g=i<=d[r]; else g=i<d[r]; res+=get(bg,l-1,r+1,tag&&(i==lim),g); } return tag?res:dp[bg][l][tag][ok]=res; } ll cal(ll x) { if(x<0) return 0LL;if(x==0) return 1LL; ll res=1; cnt=0; while(x) d[++cnt]=x%10,x/=10; rep(i,1,cnt) res+=get(i,i,1,i==cnt,true); return res; } int main() { int T,C=0; ll L,R; scanf("%d",&T); while(T--){ scanf("%lld%lld",&L,&R); if(L>R) swap(L,R); printf("Case %d: %lld\n",++C,cal(R)-cal(L-1)); } return 0; }
有部分数组没有必要：
#include<bits/stdc++.h> #define ll long long #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; ll dp[20][20]; int d[20],cnt; //tag维护前缀是否小于，ok维护后缀是否小于。维护二者不一样。 ll get(int bg,int l,int r,int tag,bool ok) { if(r>l) return !tag||(tag&&ok); if(!tag&&dp[bg][l]) return dp[bg][l]; int lim=tag?d[l]:9; ll res=0; rep(i,0,lim){ if(bg==l&&i==0) continue; bool g=ok; if(ok) g=i<=d[r]; else g=i<d[r]; res+=get(bg,l-1,r+1,tag&&(i==lim),g); } return tag?res:dp[bg][l]=res; } ll cal(ll x) { if(x<0) return 0LL;if(x==0) return 1LL; ll res=1; cnt=0; while(x) d[++cnt]=x%10,x/=10; rep(i,1,cnt) res+=get(i,i,1,i==cnt,true); return res; } int main() { int T,C=0;ll L,R; scanf("%d",&T); while(T--){ scanf("%lld%lld",&L,&R); if(L>R) swap(L,R); printf("Case %d: %lld\n",++C,cal(R)-cal(L-1)); } return 0; }
LightOJ - 1205：Palindromic Numbers （数位DP&回文串）
标签：i++ open play pac case out continue isp mic
原文地址：https://www.cnblogs.com/hua-dong/p/9980178.html
