题意：定义f(n,d)为数码d在1到n中出现的次数，其中d=0..9
如果f(d,k)=k，则称k是d好数
给定x和d，求不大于x的最大的d好数
x<=1e18
思路：考虑f的增长率主要和位数有关
各种位数，上限，下限剪枝……
事实上所有d好数不会超过1e11
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef pair<int,int> PII; 7 typedef pair<ll,ll> Pll; 8 typedef vector<int> VI; 9 typedef vector<PII> VII; 10 #define N 210000 11 #define M 4100000 12 #define fi first 13 #define se second 14 #define MP make_pair 15 #define pi acos(-1) 16 #define mem(a,b) memset(a,b,sizeof(a)) 17 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 18 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 19 #define lowbit(x) x&(-x) 20 #define Rand (rand()*(1<<16)+rand()) 21 #define id(x) ((x)<=B?(x):m-n/(x)+1) 22 #define ls p<<1 23 #define rs p<<1|1 24 25 const ll MOD=1e9+7,inv2=(MOD+1)/2; 26 double eps=1e-6; 27 ll INF=1e18; 28 int dx[4]={-1,1,0,0}; 29 int dy[4]={0,0,-1,1}; 30 31 ll a[N],b[N],ans,n,k,mi[N]; 32 int len; 33 34 ll read() 35 { 36 ll v=0,f=1; 37 char c=getchar(); 38 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 39 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 40 return v*f; 41 } 42 43 ll calc(ll n) 44 { 45 if(n<0) return 0; 46 int len=0; 47 while(n) 48 { 49 b[++len]=n%10; 50 n/=10; 51 } 52 ll t=0,re=0; 53 per(i,len,1) 54 { 55 rep(j,0,b[i]-1) 56 { 57 if(i>=2) re+=mi[i-2]*(i-1); 58 if(j==k) re+=mi[i-1]; 59 re+=mi[i-1]*t; 60 } 61 if(b[i]==k) t++; 62 } 63 return re+t; 64 } 65 66 void dfs(int u,ll now,ll now_,int flag,int s) 67 { 68 if(u==0) 69 { 70 if(calc(now)==now) 71 { 72 ans=now; 73 return; 74 } 75 } 76 if(s>=1) 77 { 78 if(calc(now)-now<=0&&calc(now_)-now_>=0) 79 { 80 ll l=now,r=now_,last=now; 81 while(l<=r) 82 { 83 ll mid=(l+r)>>1; 84 if(calc(mid)-mid<=0){last=mid; l=mid+1;} 85 else r=mid-1; 86 } 87 if(calc(last)-last==0) ans=max(ans,last); 88 } 89 return; 90 } 91 ll x=calc(now)-now,y=calc(now_)-calc(now),z=now-now_; 92 if(x<0&&x+y<0) return; 93 if(x>0&&x+z>0) return; 94 if(flag) 95 { 96 per(i,9,0) 97 { 98 dfs(u-1,now+mi[u-1]*i,now+mi[u-1]*(i+1)-1,1,s+(i==k)); 99 if(ans!=-INF) return; 100 } 101 } 102 else 103 { 104 dfs(u-1,now+mi[u-1]*a[u],now_,0,s+(a[u]==k)); 105 if(ans!=-INF) return; 106 per(i,a[u]-1,0) 107 { 108 dfs(u-1,now+mi[u-1]*i,now+mi[u-1]*(i+1)-1,1,s+(i==k)); 109 if(ans!=-INF) return; 110 } 111 } 112 113 } 114 115 int main() 116 { 117 //freopen("1.in","r",stdin); 118 //freopen("1.out","w",stdout); 119 120 int cas; 121 scanf("%d",&cas); 122 mi[0]=1; 123 rep(i,1,30) mi[i]=mi[i-1]*10; 124 while(cas--) 125 { 126 k=read(),n=read(); 127 ll n_=n; 128 if(n==0) 129 { 130 printf("0 "); 131 continue; 132 } 133 len=0; 134 while(n) 135 { 136 a[++len]=n%10; 137 n/=10; 138 } 139 ans=-INF; 140 n=n_; 141 dfs(len,0,n,0,0); 142 printf("%I64d ",ans); 143 144 145 } 146 147 return 0; 148 }