三进制搞下， 0 表示没出现过， 第i位为1 表示 i出现了奇数次， 2表示i 出现了偶数次。
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> #include<math.h> using namespace std; typedef long long LL; LL dp[20][101111]; int up[11111]; int judge(int x) { int ans=0; int a[10]; memset(a,0,sizeof(a)); while(x){ a[ans++]=x%3; x/=3; } for(int i = 0 ;i< ans;i++){ if(a[i]==1&&(i&1)) return 0 ; if(a[i]==2&&!(i&1)) return 0; } return 1; } int change(int x,int i) { int ans=0 ;int a[10]; memset(a,0,sizeof(a)); while(x){ a[ans++]=x%3; x/=3; } if(a[i]==0) a[i]=1; else if(a[i]==1) a[i]=2; else if(a[i]==2) a[i]=1; int ans1=0; for(int i = 9;i>=0;i--) ans1=ans1*3+a[i]; return ans1; } LL gao(int now,int gaojici,int first,int flag) { if(now<=0) return judge(gaojici); if(!flag&&~dp[now][gaojici]) return dp[now][gaojici]; LL limit = flag? up[now]: 9,ret=0; for(LL i= 0;i<=limit;i++){ LL kk=change(gaojici,i); ret+=gao(now-1,(first||i)?kk:0,first||i,flag&&limit==i); } return flag? ret: dp[now][gaojici]=ret; } LL solve(LL x) { int len=0; while(x){ up[++len]= x%10; x/=10; } return gao(len,0,0,1); } int main() { int Icase;LL b;LL a; memset(dp,-1,sizeof(dp)); scanf("%d",&Icase); while(Icase--){ cin>>a>>b; cout<<solve(b)-solve(a-1)<<endl; } return 0; }