The topic meaning is quite simple, look at the input data output data is more obvious
I can write it in sort or AC.
//#pragma COMMENT (linker, "/stack:16777216")//For C + + Compiler#include <stdio.h>#include<iostream>#include<fstream>#include<cstring>#include<cmath>#include<stack>#include<string>#include<map>#include<Set>#include<list>#include<queue>#include<vector>#include<algorithm>#defineMax (b) ((a) > (b))? (a): (b))#defineMin (b) ((a) < (a))? (a): (b))#defineAbs (x) (((x) > 0)? (x): (-(x)))#defineMOD 1000000007#definePi ACOs (-1.0)using namespaceStd;typedefLong Longll; typedef unsignedLong Longull; typedef unsignedint UINT; typedef unsignedCharUchar; template<classT> InlinevoidCheckmin (T &a,t b) {if(a>b) a=b;} Template<classT> InlinevoidCheckmax (T &a,t b) {if(a<b) a=b;}Const DoubleEPS = 1e-7 ;Const intN = About ;Const intM =1100011*2 ;Constll P =10000000097ll;Const intMAXN =10900000 ;Const intINF =0x3f3f3f3f ;Const intoffset = - ;inta[120000], b[120000], c[120000];intN;intMain () {Std::ios::sync_with_stdio (false); intI, J, T, K, U, V, numcase =0; CIN>>N; for(i =0; I < n; ++i) Cin >>A[i]; for(i =0; I < n-1; ++i) Cin >>B[i]; for(i =0; I < n-2; ++i) Cin >>C[i]; Sort (A, a+N); Sort (b, B+ N-1); Sort (c, C+ N-2); for(i =0; I < n-1; ++i) {if(A[i]! =B[i]) {cout<< A[i] <<Endl; Break; } } if(i = = N-1) {cout<< A[n-1] <<Endl; } for(i =0; I < n-2; ++i) {if(B[i]! =C[i]) {cout<< B[i] <<Endl; Break; } } if(i = = N-2) {cout<< B[n-2] <<Endl; } return 0;}Sort Solution
But there is a quicker way to read the official solution.
Because the size of the data in the N-10^5 ai-10 ^ 9, if all the numbers to make an accumulation of words can be stored in a long long type of variable
Source Code:
/****************************************** Solution by Bekzhan Kassenov **************************************** **/#include<bits/stdc++.h>using namespacestd;#defineF First#defineS Second#defineMP Make_pair#defineAll (x) (x). Begin (), (x). End ()typedefLong Longll;typedef unsignedLong LongUll;typedefLong Doubleld;Const DoubleEPS = 1e-9;Const DoublePI = ACOs (-1.0);Const intMOD = +* +* ++7;Const intINF = -* +* +; template<typename t>inline T Sqr (t N) {returnn *N;}intn, x;Long LongA, B, C;intMain () {#ifndef Online_judge freopen ("inch","R", stdin);#endifscanf ("%d", &N); for(inti =0; I < n; i++) {scanf ("%d", &x); A+=x; } for(inti =0; I < n-1; i++) {scanf ("%d", &x); b+=x; } for(inti =0; I < n-2; i++) {scanf ("%d", &x); C+=x; } printf ("%i64d\n%i64d\n", A-A, B-c); return 0;}
Codeforces 519B A and B and compilation Errors "Analog"