NOJ——西北工业大学C语言练习（3）_noj西工大答案

#include <stdio.h>
int main()
{
    int a = 1, b = 0, t;
    scanf("%d", &t);
    for (int i = 0; i < t;i++){
        int temp_a = b;
        int temp_b = a * 3 + 2 * b;
        a = temp_a;
        b = temp_b;
    }
    printf("%d %d", a, b);
}
1
2
3
4
5
6
7
8
9
10
11
12
13

#include <stdio.h>
int main()
{
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    for (int i = 10; i <= 100;i++){
        if(i % 3 == a && i % 5 == b && i % 7 == c){
            printf("%d", i);
            return 0;
        }
    }
    printf("-1");
}
1
2
3
4
5
6
7
8
9
10
11
12
13

#include <stdio.h>
int seven(int n){
    if (n % 7 == 0)
        return 1;
    else {
        while(n){
            if(n % 10 == 7){
                return 1;
            }
            n /= 10;
        }
        return 0;
    }
}
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 7; i <= n;i++){
        if(seven(i)){
            printf("%d ", i);
        }
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

#include <stdio.h>
#include <math.h>
typedef long long ll;

// 求x, y，使得ax + by = gcd(a, b)
ll exgcd(ll a, ll b, ll *x, ll *y)
{
    if (!b)
    {
        *x = 1;
        *y = 0;
        return a; //到达递归边界开始向上一层返回
    }
    ll d = exgcd(b, a % b, x, y);
    ll temp = *x; //推出这一层的x，y
    *x = *y;
    *y = temp - (a / b) * *y;
    return d;
}
int main()
{
    ll x, y, a, b, l;
    scanf("%lld%lld%lld%lld%lld", &x, &y, &a, &b, &l);

    ll m = 0, n = 0, o, s;
    o = exgcd(abs(a - b),l, &m, &n); // a表示跳的次数，b表示圈数
    if ((x - y) % o != 0 || a == b)
        printf("Impossible\n");
    else
    {
        s = l / o;
        m = m * ((x - y) / o);
        m = (m % s + s) % s;
        printf("%lld", m);
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36

#include <stdio.h>
int main()
{
    int a, b, n = 1;
    double m = 1;
    scanf("%d%d", &a,&b);
    double sum = 1;
    int begin = 1;
    int end = 1;
    int flag = 0;
    while(sum < b){
        n++;
        m += 1.0 / (double)n;
        sum += 1.0 / (double)m;
        if(sum > a){
            if(flag == 0){
                begin = n;
                flag = 1;
            }
            if(sum < b){
                end = n;
            }
        }
    }
    printf("%d %d", begin, end);
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26

#include <stdio.h>
int isPrime(int n)
{
    for (int i = 2; i * i <= n;i++){
        if(n % i ==0)
            return 0;
    }
    return 1;
}
int main()
{
    int n, cnt1 = 1, cnt2 = 0,sum_number;
    scanf("%d", &n);
    while(1)
    {
        sum_number = 1;
        for (int i = (cnt1 - 1) * 100 + 1; i <= (cnt1 - 1) * 100 + 99;i += 2)
        {
            if(isPrime(i)){
                sum_number = 0;
                break;
            }
        }
        if(sum_number == 1){
            cnt2++;
        }
        if(cnt2 == n){
            printf("%d %d", (cnt1 - 1) * 100 + 0, (cnt1 - 1) * 100 + 99);
            break;
        }
        cnt1++;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33

#include <stdio.h>

void recheck(int a,int b,int n)
{
    int num[10];
    for (int i = 0; i < 5; i++){
        num[i] = a % 10;
        a = a / 10;
    }

    for (int i = 0; i < 5; i++)
    {
        num[i + 5] = b % 10;
        b = b / 10;
    }
    for (int k = 0; k < 10; k++) //判断a和b的各位数字是否有重复
    {
        for (int j = 0; j < k;j++){
            if(num[k] == num[j])
                return;
        }
    }
    printf("%d%d%d%d%d/%d%d%d%d%d=%d\n", num[9], num[8], num[7], num[6], num[5], num[4], num[3], num[2], num[1], num[0], n);
    
    return; //若各位数字都不同，则返回1
}

int main()
{
    int n, a, b;
    scanf("%d", &n);
    for (int i = 1234;i * n <=98765;i++){
        recheck(i, n * i, n);
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

#include <stdio.h>
int main()
{
    int k;
    scanf("%d", &k);
    for (int i = k + 1; i <= 2 * k;i++)
    {
        if((i * k) % (i - k) == 0)
        {
            printf("1/%d=1/%d+1/%d\n", k,  (i * k) / (i - k),i);
        }
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13

#include <stdio.h>
int main()
{
    int a, b, c;
    int num[102];
    scanf("%d%d%d", &a, &b, &c);
    int ans = a / b, carry = 0;
    num[0] = ans;
    for (int i = 1; i <= c + 1;i++)
    {
        a = a * 10;
        ans = (a / b) % 10;
        a = a % b;
        num[i] = ans;
    }
    if(num[c+1] >= 5){
        num[c]++;
    }
    for (int i = c; i >= 1;i--)
    {
        if(num[i] > 9){
            num[i] -= 10;
            num[i - 1] += 1;
        }
    }
    printf("%d.",num[0]);
    for (int i = 1; i <= c;i++){
        printf("%d",num[i]);
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

#include <stdio.h>
void recheck(int a, int b, int c)
{
    int num[9];
    for (int i = 0; i < 3; i++)
    {
        num[i] = a % 10;
        a = a / 10;
    }

    for (int i = 0; i < 3; i++)
    {
        num[i + 3] = b % 10;
        b = b / 10;
    }
    for (int i = 0; i < 3; i++)
    {
        num[i + 6] = c % 10;
        c = c / 10;
    }
    for (int k = 0; k < 9; k++)
    {
        for (int j = 0; j < k; j++)
        {
            if (num[k] == num[j] || num[k] == 0)
                return;
        }
    }
    printf("%d%d%d %d%d%d %d%d%d\n", num[2], num[1], num[0], num[5], num[4], num[3], num[8], num[7], num[6]);

    return; //若各位数字都不同，则返回1
}
int main()
{
    for (int i = 111; i <= 333;i++)
    {
        int j = i * 2;
        int k = i * 3;
        recheck(i, j, k);
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41