2018年武漢大學(xué)計(jì)算機(jī)復(fù)試上機(jī)題

下面是使用Java編寫的個人答案（非參考答案，建議自己參考網(wǎng)上的C語言代碼）：

第1題：給定兩個正整數(shù)a和b爹耗，求在[a，b]中的所有整數(shù)中，每個數(shù)碼（digit 0~9）各出現(xiàn)了多少次簿煌。
例如輸入：
1 99
輸出
9 20 20 20 20 20 20 20 20 20

public class test1 {
    public static void main(String[] args) {
        String a, b;
        int[] num = new int[10];
        for (int i = 0; i < 10; i++) {
            num[i] = 0;
        }
        Scanner in = new Scanner(System.in);
        a = in.next();
        b = in.next();
        BigInteger ansA = new BigInteger(a);
        BigInteger ansB = new BigInteger(b);
        //考慮從a到b-1的所有數(shù)
        while (!ansA.equals(ansB)) {
            char[] compare = ansA.toString().toCharArray();
            for (int i = 0; i < ansA.toString().length(); i++) {
                for (int j = 0; j < 10; j++) {
                    if (compare[i] - '0' == j) {
                        num[j]++;
                    }
                }
            }
            ansA = ansA.add(new BigInteger("1"));
        }
        //單獨(dú)考慮b
        for (int i = 0; i < ansB.toString().length(); i++) {
            char[] compare = ansB.toString().toCharArray();
            for (int j = 0; j < 10; j++) {
                if (compare[i] - '0' == j) {
                    num[j]++;
                }
            }
        }
        for (int i = 0; i < 10; i++) {
            System.out.println(num[i]);
        }
    }
}

C代碼：(不考慮大整數(shù)...)

int main() {
    int hashTable[10] = {0};
    int a, b;
    while (scanf("%d %d", &a, &b) != EOF) {
        for (int i = a; i <= b; i++) {
            int n = i;
            while (n > 0) {
                int x = n % 10;
                hashTable[x]++;
                n = n / 10;
            }
        }
        for (int i = 0; i < 10; i++) {
            printf("%d ", hashTable[i]);
        }
        printf("\n");
    }
    return 0;
}

第2題：輸入一個整數(shù)n（0<n<10），顯示n行如下規(guī)律圖形鉴吹。
例如輸入3 姨伟，顯示
1
2 3
4 5 6
例如輸入5，顯示
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15

public class test2 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int count = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n - i - 1; j++) {
                System.out.print("   ");
            }
            for (int j = n - i - 1; j < n; j++) {
                System.out.printf("%3d", count);
                count++;
            }
            System.out.println();
        }
    }
}

C代碼

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;

int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
        int x = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n - i - 1; j++) {
                printf("   ");
            }
            for (int j = n - i - 1; j < n; j++) {
                printf("%3d", x);
                x++;
            }
            printf("\n");
        }
    }
    return 0;
}

第3題:因?yàn)?51既是一個質(zhì)數(shù)又是一個回文數(shù)（從左到右和從右到左看是一樣的）豆励，所以151是回文質(zhì)數(shù)夺荒。
寫一個程序來找出范圍[a,b]間的所有回文質(zhì)數(shù)，a,b由鍵盤輸入良蒸。(這題眼熟嗎技扼？對啊诚啃！因?yàn)橛钟谐笳麛?shù)這個坑)

public class test3 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String a = in.next();
        String b = in.next();
        in.close();
        BigInteger ansA = new BigInteger(a);
        BigInteger ansB = new BigInteger(b);
        while (!ansA.equals(ansB)) {
            if (isPrimeNumber(ansA)) {
                char count[] = ansA.toString().toCharArray();
                int time = 0;
                for (int i = 0; i < ansA.toString().length() / 2; i++) {
                    if (count[i] == count[count.length - i - 1]) {
                        time++;
                    }
                }
                if (time == ansA.toString().length() / 2) {
                    System.out.println(ansA.toString());
                }
            }
            ansA = ansA.add(new BigInteger("1"));
        }
    }

    private static boolean isPrimeNumber (BigInteger num) {
        BigInteger i = new BigInteger("2");
        while (i.compareTo(num) <= 0) {
            if ((num.mod(i)).equals(new BigInteger("0"))) {
                break;
            }
            i = i.add(new BigInteger("1"));
        }
        return i.equals(num);
    }
}

第4題：輸入一個N（N<=10）階方陣淮摔，按照如下方式調(diào)整方陣：
1.將第一列中最大數(shù)所在的行與第一行對調(diào)。
2.將第二列中從第二行到第N行最大數(shù)所在的行與第二行對調(diào)始赎。
依此類推...
N-1.將第N-1列中從第N-1行到第N行最大數(shù)所在的行與第N-1行對調(diào)和橙。
N.輸出這個方陣
輸入：
包含多組測試數(shù)據(jù),每組測試數(shù)據(jù)第一行為一個整數(shù)N,表示方陣的階數(shù).
接下來輸入這個N階方陣.
輸出：
調(diào)整后的方陣
樣例輸入：
4
3 6 8 7
6 7 5 3
8 6 5 3
9 8 7 2
樣例輸出：
9 8 7 2
6 7 5 3
3 6 8 7
8 6 5 3

public class test4 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int num[][] = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                num[i][j] = in.nextInt();
            }
        }
        for (int i = 0; i < n; i++) {
            int max = i;
            for (int j = i; j < n; j++) {
                if (num[j][i] > num[max][i]) {
                    max = j;
                }
            }
            int temp[] = new int[n];
            for (int j = 0; j < n; j++) {
                temp[j] = num[i][j];
                num[i][j] = num[max][j];
                num[max][j] = temp[j];
            }

        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                System.out.print(num[i][j] + " ");
            }
            System.out.println();
        }
    }
}

第5題：兩個數(shù)字字符串相加
有定義：char s1[200]，s2[200]造垛，s3[200]
若輸入s1和s2非全數(shù)字字符串魔招，顯示輸入錯誤；
否則計(jì)算s1與s2相加后的結(jié)果五辽，存放于s3并顯示办斑。
例如輸入：
999999999999999999999
999999999999999999999
例如輸出
1999999999999999999998
輸入：
1234567890123456789
876543211
輸出：
1234567891000000000

public class test5 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s1 = in.nextLine();
        String s2 = in.nextLine();
        char ansA[] = s1.toCharArray();
        char ansB[] = s2.toCharArray();
        for (int i = 0; i < s1.length(); i++) {
            if (ansA[i] < '0' || ansA[i] > '9') {
                System.out.println("輸入錯誤");
                System.exit(0);
            }
        }
        for (int i = 0; i < s2.length(); i++) {
            if (ansB[i] < '0' || ansB[i] > '9') {
                System.out.println("輸入錯誤");
                System.exit(0);
            }
        }
        BigInteger a = new BigInteger(s1);
        BigInteger b = new BigInteger(s2);
        System.out.println(a.add(b));
    }
}

C代碼：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;

struct bign {
    int d[1010];
    int len;
    bign () {
        memset(d, 0, sizeof(d));
    }
};

bign change(char str[]) {
    bign a;
    a.len = strlen(str);
    for (int i = 0; i < a.len; i++) {
        a.d[i] = str[a.len - i - 1] - '0';
    }
    return a;
}

bign add(bign a, bign b) {
    bign c;
    int carry = 0;
    for (int i = 0; i < a.len || i < b.len; i++) {
        int temp = a.d[i] + b.d[i] + carry;
        c.d[c.len++] = temp % 10;
        carry = temp / 10;
    }
    if (carry != 0) {
        c.d[c.len++] = carry;
    }
    return c;
}
int main() {
    char str1[200], str2[200];
    bool flag = true;
    while(scanf("%s %s", str1, str2) != EOF) {
        int len1 = strlen(str1), len2 = strlen(str2);
        for (int i = 0; i < len1; i++) {
            if (str1[i] > '9' || str1[i] < '0') {
                flag = false;
            }
        }
        for (int i = 0; i < len2; i++) {
            if (str2[i] > '9' || str2[i] < '0') {
                flag = false;
            }
        }
        if (flag == false) {
                printf("input error!\n");
            continue;
        }
        bign a = change(str1);
        bign b = change(str2);
        bign c = add(a, b);
        for (int i = c.len - 1; i >= 0; i--) {
            printf("%d", c.d[i]);
        }
        printf("\n");
    }
    return 0;
}