OJ 2018/3/9

我会带着你远行 2022-05-30 10:10 95阅读 0赞

/* A1012. 进制转换3 long long 使用 _int64 FFFFFFFF 8个F 是最大边界 0是 最小边界 这里有个特别注意 VC++6.0 可以定义 _int64 替代 long long 但是 OJ 只认 long long */
    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    int main()
    {
        int i;
        char th[8];
        gets(th);   
        int num=strlen(th);
        if(num>8||num==0||th[0]=='-')
            return 0;
        _int64 total=0;
        for(i=0;i<num;i++)
        {   
            int temp=0;
            if('0'<=th[num-i-1] && th[num-i-1]<='9')
            {
                temp=(int)th[num-i-1]-48;
                total+=(temp*pow(16,i));
    
            }else if('A'<=th[num-i-1] && th[num-i-1]<='F')
            {
                temp=(int)th[num-i-1]-55;
                total+=(temp*pow(16,i));
            }   
            else
                return 0;
        }   
        printf("%I64d",total);
        return 0;
    }

/* A1012. 进制转换5 long long 使用 _int64 FFFFFFFF 8个F 是最大边界 0是 最小边界 */
    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    void f(int len,char b[100001])
    {
        long long total=0;
        int i;
        for(i=0;i<len;i++)
        {
            int temp;
            if(b[len-i-1]>='0' && b[len-i-1]<='9')
                temp=b[len-i-1]-48;
            else if(b[len-i-1]>='A' && b[len-i-1]<='F')
                temp=b[len-i-1]-55;
            total+=temp*pow(16,i);          
        }   
        long long t[100];
        long long chu,yu;
        long long flag=0;
        while(total)
        {
            chu=total/8;
            yu=total%8;
            t[flag]=yu;
            total /= 8;
            flag++;
        }
        for(i=flag-1;i>=0;i--)
            printf("%d",t[i]);
    
    }
    int main()
    {
        int i;
        int num;
        scanf("%d",&num);
        if(num<1||num>10)
            return 0;
        char a[10][100001];
        int b[10];
            for(i=0;i<num;i++)
            scanf("%s",&a[i]);  
        for(i=0;i<num;i++)
        {
            b[i]=strlen(a[i]);
            if(b[i]>100000)
                return 0;
        }
    
    
        for(i=0;i<num;i++)
        {
            f(b[i],a[i]);
            printf("\n");
        }       
        return 0;
    }

/* AC */
    #include <stdio.h>
    #include <string.h>
    
    int main(int argc, char *argv[])
    {
        char *s1[16] = {
       "0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111",
                        "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"
                       };
        register int i, beg, j, k;
        int n;
        char input[100100], htb[400102];
        int resu[133434], len;
        scanf("%d", &n);
    
        while (n--)
        {
            scanf("%s", input);
    
            len = strlen(input);
            if (len % 3 == 0)
                beg = 0;
            else if (len % 3 == 1)
            {
                beg = 2;
                htb[0] = htb[1] = '0';
            }
            else
                {
                    beg = 1;
                    htb[0] = '0';
                }
    
                if (len * 4 % 3 == 0)
                    k = len * 4;
                else if ((len * 4 + 1) % 3 == 0)
                    k = len * 4 + 1;
                else
                    k = len * 4 + 2;
    
                for (i = 0; i <= len - 1; i++)
                {
                    if (input[i] <= 57)
                    {
                        htb[beg] = s1[(int)(input[i] - 48)][0];
                        htb[beg + 1] = s1[(int)(input[i] - 48)][1];
                        htb[beg + 2] = s1[(int)(input[i] - 48)][2];
                        htb[beg + 3] = s1[(int)(input[i] - 48)][3];
                        beg += 4;
                    }
                    else
                    {
                        htb[beg] = s1[(int)(input[i] - 55)][0];
                        htb[beg + 1] = s1[(int)(input[i] - 55)][1];
                        htb[beg + 2] = s1[(int)(input[i] - 55)][2];
                        htb[beg + 3] = s1[(int)(input[i] - 55)][3];
                        beg += 4;
                    }
                }
    
                resu[0] = (int)(htb[0] - 48) * 4 + (int)(htb[1] - 48) * 2 + (int)(htb[2] - 48);
                for (j = 1; j <= k / 3 - 1; j++)
                {
                    resu[j] = (int)(htb[j * 3] - 48) * 4 + (int)(htb[j * 3 + 1] - 48) * 2 +
                              (int)(htb[j * 3 + 2] - 48);
                }
    
                if (resu[0] != 0)
                    printf("%d", resu[0]);
    
                for (j = 1; j <= k / 3 - 1; j++)
                    printf("%d", resu[j]);
                printf("\n");
            }
    
            return 0;
        }

/*
    杨辉三角
    */
    #include<stdio.h>
    int main()
    {
        int a[(1+34)*2][(1+34)*2]={1};
        int i,j;
        int n;
        scanf("%d",&n);
        if(n<1||n>34)
            return 0;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=i;j++)
            {
                a[i][j]=a[i-1][j]+a[i-1][j-1];
                printf("%d ",a[i][j]);
            }
            printf("\n");
        }
        return 0;
    }