CodeForces 519B A and B and Compilation Errors【模拟】

题目意思还是蛮简单的，看 输入数据输出数据还是比较明显的

我用排序来写还是可以AC的

//#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>

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

#define MOD 1000000007

#define pi acos(-1.0)



using namespace std;



typedef long long           ll      ;

typedef unsigned long long  ull     ;

typedef unsigned int        uint    ;

typedef unsigned char       uchar   ;



template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}

template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}



const double eps = 1e-7      ;

const int N = 210            ;

const int M = 1100011*2      ;

const ll P = 10000000097ll   ;

const int MAXN = 10900000    ;

const int INF = 0x3f3f3f3f   ;

const int offset = 100       ;



int a[120000], b[120000], c[120000];

int n;



int main () {

    std::ios::sync_with_stdio (false);

    int i, 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

不过看了官方题解，还有更快的方法。

因为这题的数据规模在 n - 10^5  ai - 10 ^ 9 ，如果对所有的数字作一个累加的话还是可以存放在 long long 类型的变量中

 

Source Code:

/****************************************

**     Solution by Bekzhan Kassenov    **

****************************************/



#include <bits/stdc++.h>



using namespace std;



#define F first

#define S second

#define MP make_pair

#define all(x) (x).begin(), (x).end()



typedef long long ll;

typedef unsigned long long ull;

typedef long double ld;



const double EPS = 1e-9;

const double PI = acos(-1.0);

const int MOD = 1000 * 1000 * 1000 + 7;

const int INF = 2000 * 1000 * 1000;



template <typename T>

inline T sqr(T n) {

    return n * n;

}



int n, x;

long long a, b, c;



int main() {

#ifndef ONLINE_JUDGE

    freopen("in", "r", stdin);

#endif



    scanf("%d", &n);



    for (int i = 0; i < n; i++) {

        scanf("%d", &x);

        a += x;

    }



    for (int i = 0; i < n - 1; i++) {

        scanf("%d", &x);

        b += x;

    }



    for (int i = 0; i < n - 2; i++) {

        scanf("%d", &x);

        c += x;

    }



    printf("%I64d\n%I64d\n", a - b, b - c);



    return 0;

}