MS 实习生热身赛

题目1 : Arithmetic Expression

描述

Given N arithmetic expressions, can you tell whose result is closest to 9?

输入

Line 1: N (1 <= N <= 50000).
Line 2..N+1: Each line contains an expression in the format of "a op b" where a, b are integers (-10000 <= a, b <= 10000) and op is one of addition (+), subtraction (-), multiplication (*) and division (/). There is no "divided by zero" expression.

输出

The index of expression whose result is closest to 9. If there are more than one such expressions, output the smallest index.

样例输入

4

901 / 100

3 * 3

2 + 6

8 - -1

样例输出

2

 

解法:

1. 直接上 double 比较, 只过了部分案例(不排除代码本身的 bug)

2. 这道题考察的应该是分数的表示, 考试的时候我大致比划了一下, 但没有实现, 考试之后提交系统已经关闭, 因此也不知后面写出的代码是否有问题.

唯有 / 会带来小数的问题. 我们用到分数的一个性质 a/b > c/d <-> a*d > b*c.

因此我们将所有的数都表示成 abs(divided/divisor -9) 的形式.

比如 901 / 100 -> 1/100; 3*3 -> 0/1; 2+6 -> 1/1; 8 - -1 = 0/1;

这样就不会有小数的问题了.

#include <iostream>

#include <stdio.h>

#include <memory.h>

#include <algorithm>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <deque>

#define MIN(x,y) (x)<(y)?(x):(y)

using namespace std;

 

int main() {

    freopen("C:\Users\vincent\Dropbox\workplacce\joj\test.txt", "r", stdin);

 

    int T, iCase = 0;

    scanf("%d", &T);

    

    char op1[1000], op2[1000];

    char ops[10];

    

    int divisor = 1, divided = 9999999;

    int index = 0;

 

    while(T --) {

        iCase ++;

        scanf("%s%s%s", op1,ops, op2);

        int lhs = atoi(op1);

        int rhs = atoi(op2);

        char op = ops[0];

        int fenzi, fenmu;

 

        if(op == '+') {

            fenzi = lhs + rhs;

            fenmu = 1;

        }else if(op == '-') {

            fenzi = lhs - rhs;

            fenmu = 1;

        }else if(op == '*') {

            fenzi = lhs * rhs;

            fenmu = 1;

        }else {

            fenzi = lhs;

            fenmu = rhs;

        }

 

        fenzi = fenzi - fenmu * 9;

 

        if(fenzi < 0) {

            fenzi = -1*fenzi;

        }

        if(fenmu < 0) {

            fenmu = -1 * fenmu;

        }

 

        if(divided*fenmu > divisor*fenzi) {

            index = iCase;

            divided = fenzi;

            divisor = fenmu;

        }

    }

    printf("%d ", index);

    return 0;

}

 

题目2 : Longest Repeated Sequence

描述

You are given a sequence of integers, A = a₁, a₂, ... a_n. A consecutive subsequence of A (say a_i, a_i+1 ... a_j) is called a "repeated sequence" if it appears more than once in A (there exists some positive k that a_i+k = a_i, a_i+k+1 = a_i+1, ... a_j+k = a_j) and its appearances are not intersected (i + k > j).

Can you find the longest repeated sequence in A?

输入

Line 1: n (1 <= n <= 300), the length of A.
Line 2: the sequence, a₁ a₂ ... a_n (0 <= a_i <= 100).

输出

The length of the longest repeated sequence.

样例输入

5

2 3 2 3 2

样例输出

2

 

思路:

DP 题目. dp[i][j] 表示以 i 为结尾和以 j 为结尾相同子序列的长度

dp[i][j] = dp[i-1][j-1] + 1 if(arr[i] == arr[j], 且两段子序列不相交)

 

#include <iostream>

#include <stdio.h>

#include <memory.h>

#include <algorithm>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <deque>

#define MIN(x,y) (x)<(y)?(x):(y)

using namespace std;

 

int arr[500];

int dp[500][500];

 

int solve_dp(int n) {

    memset(dp, 0, sizeof(dp));

 

    int maxval = 0;

 

    for(int i = 1; i <= n; i ++) {

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

 

            if(arr[i] == arr[j]) {

                if(i + dp[i-1][j-1] < j) {

                    dp[i][j] = dp[i-1][j-1] + 1;

                }else {

                    dp[i][j] = dp[i-1][j-1];

                }

            }

            maxval = max(maxval, dp[i][j]);

 

        }

    }

    return maxval;

}

 

int main() {

    freopen("C:\Users\vincent\Dropbox\workplacce\joj\test.txt", "r", stdin);

 

    int n;

    scanf("%d", &n);

 

    for(int i = 1; i <= n; i ++) {

        scanf("%d", arr+i);

    }

    

    cout << solve_dp(n) << endl;

 

    return 0;

}