Test instructions: A city citizen each has a five-digit ID, the first four bits is the basic ID, the last one is the check digit, the check digit is based on the sample given table, the ID of a citizen is ABCDE, then E= ((((0$a) $b) $d), every citizen can make mistakes, For example, if one digit is wrong or the adjacent two is swapped, then our excellent table should be able to judge the situation of error, we know that for the correct ID, (((((((((((() $b) $d) $e) = 0, but equal to 0 is not necessarily correct, not equal to 0 must be wrong, We are asking for the four-bit basic ID we can not verify that this is the correct number of IDs, that is, as long as this ID error condition may be sentenced to correct, then it is not possible, ans will be Gaga
Practice: Direct Simulation
Pit point: For example, 1233 33 after the exchange with the original value return value is the same, so even if the same can not be judged correct, because the 33 exchange is the original value, here special consideration can be.
The code is as follows:
#include <stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespacestd;intmp[Ten][Ten];intnum[Ten];intans;//int vis[10];intsolve_1 () {intres =0; for(intI=0; i<4; i++) {res=Mp[res][num[i]]; } returnRes;}intsolve () {intres =0; for(intI=0; i<=4; i++) {res=Mp[res][num[i]]; } returnRes;}BOOLok () {//memset (Vis, 0, sizeof (VIS)); intttmp; for(intI=0; i<5; i++) {ttmp=Num[i]; for(intj=0; j<Ten; J + +) { if(j==ttmp)Continue; Num[i]=J; if(Solve () = =0) return false; } Num[i]=ttmp; } if(num[0]! = num[1]) {Swap (num[0], num[1]); if(Solve () = =0) return false; Swap (num[0], num[1]); } if(num[1]! = num[2]) {Swap (num[1], num[2]); if(Solve () = =0) return false; Swap (num[1], num[2]); } if(num[2]! = num[3]) {Swap (num[2], num[3]); if(Solve () = =0) return false; Swap (num[2], num[3]); } if(num[3]! = num[4]) {Swap (num[3], num[4]); if(Solve () = =0) return false; } return true;}intMain () {ans=0; for(intI=0; i<Ten; i++) for(intj=0; j<Ten; J + +) scanf ("%d", &Mp[i][j]); for(intI=0; i<Ten; i++) { for(intj=0; j<Ten; J + +) { for(intp=0; p<Ten; p++) { for(intq=0; q<Ten; q++) {num[0] =i; num[1] =J; num[2] =p; num[3] =Q; num[4] =solve_1 (); if( !OK ()) {ans++;//printf ("%d%d%d%d \ n", Num[0], num[1], num[2], num[3], num[4]);}}}}} printf ("%d\n", ans); return 0;}
The Asia regional Contest, Tsukuba quality of Check Digits gym-101158b