蓝桥练习部分题

5顺时针螺旋数：

#include<stdio.h>
#include<string.h>

int map[25][25];

int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        memset(map, 0, sizeof(25));
        int temp = 1;
        int x , y;
        int up = 1, down = n;
        int left = 1, right = n;
        while(temp <= n*n)
        {
            for(int i = left; i <= right; i++)
            {
                map[up][i] = temp++;
            }
            up++;

            for(int i = up; i <= down; i++)
            {
                map[i][right] = temp++;
            }
            right--;

            for(int i = right; i >= left; i--)
            {
                map[down][i] = temp++;
            }
            down--;

            for(int i = down; i >= up; i--)
            {
                map[i][left] = temp++;
            }
            left++;
        }

        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                printf("%d ", map[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}

6.从键盘输入一个日期，格式 yyyy-M-d,要求计算该日期与 1949 年 10月 1 日距离多少天

例如 ：

1949-10-2

1

1949-11-1

2

#include<stdio.h>
int ms[13] = {0,31,28,31,30,31,30,31,30,31,31,30,31,30,31};
int main()
{
    int year, month, day;
    char c1, c2;
    while(scanf("%d%c%d%c%d",&year, &c1, &month, &c2, &day) != EOF)
    {
        int ans = 0;

        if(y == 1949)
        {
            for(int i = 10; i < month; i++)
                ans += ms[i];
            ans += day;
            ans--;
        }
        else
        {
            ans += ms[10]+ms[11]+ms[12]-1; //1949
            for(int y = 1950; y < year; y++) //1949----year-1
            {
                if((y%4==0 && y%100 != 0) || y%400 == 0) ans += 366; //ÈóÄê
                else ans += 365; //ƽÄê
            }

            for(int m = 1; m < month; m++) //1----month-1
            {
                ans += ms[i];
            }

            ans += day;
            if((year%4==0 && year%100 !=0) || year%400==0)
            {
                if(month > 2) ans++;
            }


        }
        printf("%d\n", ans);

    }
    return 0;
}

7信用卡卡号检测

#include<stdio.h>
#include<string.h>

int a[100];
char str[100];
int main()
{
    while(scanf("%s", str))
    {
        int len = strlen(str);
        int j = 1;
        for(int i = len-1; i >= 0; i--)
        {
            a[j++] = str[i]-'0';
        }

        int odd = 0; //奇数
        int even = 0;
        for(int i = 1; i <= len; i++)
        {
            if(i%2 == 0)
            {
                even += a[i]*2 >= 10 ? a[i]*2-9 : a[i]*2;
            }
            else odd += a[i];
        }

        if((odd+even) % 10 == 0) printf("成功\n");
        else printf("失败\n");
    }
    return 0;
}

13平方的末位为本身【效率较低】

#include<stdio.h>
int main()
{
    for(int i = 100; i < 10000; i++)
    {
        if(i*i%1000 == i) printf("%d\n", i);
    }
    return 0;
}

14.概率【结果好像有点问题】

#include<stdio.h>
#include<math.h>
int main()
{
    int total = 2*2*2*2;
    double n = 0;
    double a = 1000;
    double b = 2000;
    for(int i = 0; i <=1; i++)
    {
        if(i == 0)
        {
            a += b/4;
            b = b/4*3;
        }
        else
        {
            b += a/4;
            a = a/4*3;
        }
        for(int j = 0; j <= 1; j++)
        {
            if(j == 0)
            {
                a += b/4;
                b = b/4*3;
            }
            else
            {
                b += a/4;
                a = a/4*3;
            }

            for(int k = 0; k <= 1; k++)
            {
                if(k == 0)
                {
                    a += b/4;
                    b = b/4*3;
                }
                else
                {
                    b += a/4;
                    a = a/4*3;
                }
                for(int e = 0; e <= 1; e++)
                {
                    if(e == 0)
                    {
                        a += b/4;
                        b = b/4*3;
                    }
                    else
                    {
                        b += a/4;
                        a = a/4*3;
                    }
                    if((b-a<1000 && b-a>0) || (a-b<1000 && a-b>0)) n++;
                    //printf("%d %d %d %d %lf %lf\n", i, j, k, e, a, b);
                    //printf("%lf \n", a-b);
                }
            }
        }
    }
    //printf("%lf\n", n);
    printf("%.3lf\n",(double)n/16.0);
    return 0;
}

15打印整数的划分

#include<stdio.h>

int n, cnt;
int a[21];

void fun(int divi, int sum)
{
    if(sum == n) //注意最后一位的输出
    {
        for(int i = 0; i < cnt-1; i++)
            printf("%d+", a[i]);
        if(n-a[0] == cnt-1) printf("%d\n", a[cnt-1]);
        else printf("%d,", a[cnt-1]);
        return;
    }

    for(int i = divi; i > 0; i--) //由大到小划分
    {
        if(sum+i <= n)
        {
            a[cnt++] = i;
            fun(i, sum+i); //划分 i,而不是 n-i，因为打印的结果是由大到小排序的
                           //所以每次枚举的最大数不能超过上次的最大值,同时也避免了重复
            cnt--; //回溯,划分比i小1的
        }
    }
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        cnt = 0;
        fun(n, 0);
    }
    return 0;
}

/*
//递归种数
#include<stdio.h>

int q(int n, int m)
{
    if(n < 1 || m < 1) return 0;
    if(n==1 || m==1)
        return 1;
    if(n < m)
        return q(n, n);
    if(n == m)
        return q(n, m-1)+1;
    return q(n, m-1)+q(n-m,m);
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        int ans = q(n,n);
        printf("%d\n", ans);
    }
    return 0;
}
*/

16.21位花朵数，不会做点击打开链接