Hangzhou Computer Program Design Competition (2016 ' 12)-Network synchronization 1002

Increment number

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 0 Accepted Submission (s): 0

problem DescriptionIf a positive integer a, adjacent bits always satisfy the lower than equal to the high, it is called the increment number.
For example: 1223,667 is the increment number.
now there is a positive integer X, how many positive integer a satisfies 1<=a<=x, and a is incremented. InputThe first line of input data is a positive integer T (1<=t<=20) that represents the number of groups of test data.
Next T line, one positive integer X (1<=x<=100000000) per line. Outputfor each set of data, output a row of answers, see the output sample. Sample Input35 - - Sample Output5 - Wu

/** * @Author: H s-j * @DateTime: 2016-12-24 13:06:57 * @Description: Hdu-2*/#include<functional>#include<algorithm>#include<iostream>#include<fstream>#include<sstream>#include<iomanip>#include<numeric>#include<cstring>#include<cassert>#include<cstdio>#include<string>#include<vector>#include<bitset>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<list>#include<Set>#include<map>using namespacestd;Long Longdp[Ten][Ten] = {0};intdig[111];intCountLong Longm) {intnum =1;  while(M/Ten) {        ++num; M/=Ten; }    returnnum;}voidCount_1 (Long LongMintflag) {     while(m) {Dig[flag]= (m%Ten); ++Flag; M/=Ten; }}    voidinit () { for(inti =1; I <Ten; ++i) { for(intj =0; J <Ten; ++j) {if(i = =1) {Dp[i][j]=J; }            Else {                if(J = =0) Dp[i][j] = (Dp[i-1][9]); ElseDP[I][J] = (Dp[i][j-1] + dp[i-1][9]-Dp[i-1][j-1]); }        }    }}intMain () {init (); intN; CIN>>N;  while(n--) {        Long Longx; CIN>>x; intnum; Num=count (x); Count_1 (x,1); Long LongAns =0;  for(inti = num; i >0; --i) {if(I < num && Dig[i] < Dig[i +1]) Break; if(i = =1) {                if(num = =1) ans + = dig[1]; ElseAns + = (Dig[i]-dig[i +1] +1); }            Else {                if(i = =num) {ans+ = Dp[i][dig[i]-1]; }                Else{ans+ = (Dp[i][dig[i]-1]-Dp[i][dig[i +1] -1]); } }} cout<< ans <<Endl; }    return 0;}

Hangzhou Computer Program Design Competition (2016 ' 12)-Network synchronization 1002