题意：
1≤T≤200
● 1≤length(N)≤100
思路：
设f[i,j,k,l]为第i位为j，前i位是否贴上限(0/1)，递减或递增(0/1)方案数
g[i,j,k]为不到n位，第i位为j，递减或递增方案数
1 const mo=1000000007; 2 var f:array[0..105,0..10,0..2,0..2]of longint; 3 g:array[0..105,0..10,0..2]of longint; 4 a:array[1..105]of longint; 5 v,n,i,j,k,l,jj,kk,ans,cas,ll:longint; 6 ch:string; 7 8 begin 9 10 readln(cas); 11 for v:=1 to cas do 12 begin 13 readln(ch); 14 n:=length(ch); 15 for i:=1 to n do a[i]:=0; 16 for i:=1 to n do a[i]:=ord(ch[i])-ord('0'); 17 for i:=1 to n do 18 for j:=0 to 9 do 19 for k:=0 to 1 do 20 for l:=0 to 1 do f[i,j,k,l]:=0; 21 for i:=1 to a[1]-1 do f[1,i,0,0]:=1; 22 f[1,a[1],1,0]:=1; 23 for i:=2 to n do 24 for j:=0 to 9 do 25 for k:=0 to 1 do 26 for l:=0 to 1 do 27 for jj:=0 to 9 do 28 begin 29 if k=0 then kk:=0; 30 if (k=1)and(jj<a[i]) then kk:=0; 31 if (k=1)and(jj=a[i]) then kk:=1; 32 if (k=1)and(jj>a[i]) then break; 33 for ll:=l to 1 do 34 begin 35 if (ll=0)and(j<jj) then continue; 36 if (ll=1)and(j>jj) then continue; 37 if (ll<>l)and(j=jj) then continue; 38 f[i,jj,kk,ll]:=(f[i,jj,kk,ll]+f[i-1,j,k,l]) mod mo; 39 end; 40 end; 41 for i:=1 to n-1 do 42 for j:=0 to 9 do 43 for k:=0 to 1 do g[i,j,k]:=0; 44 for i:=1 to 9 do g[1,i,0]:=1; 45 for i:=2 to n-1 do 46 for j:=0 to 9 do 47 for k:=0 to 1 do 48 for jj:=0 to 9 do 49 for kk:=k to 1 do 50 begin 51 if (kk=0)and(j<jj) then continue; 52 if (kk=1)and(j>jj) then continue; 53 if (k<>kk)and(j=jj) then continue; 54 g[i,jj,kk]:=(g[i,jj,kk]+g[i-1,j,k]) mod mo; 55 end; 56 ans:=0; 57 for j:=0 to 9 do 58 for k:=0 to 1 do 59 for l:=0 to 1 do ans:=(ans+f[n,j,k,l]) mod mo; 60 for i:=1 to n-1 do 61 for j:=0 to 9 do 62 for k:=0 to 1 do ans:=(ans+g[i,j,k]) mod mo; 63 writeln(ans); 64 end; 65 66 end.