字符串处理

大位数乘法乘法

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!
Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.
Input Specification:
Each input contains one test case. Each case contains one positive integer with no more than 20 digits.
Output Specification:
For each test case, first print in a line “Yes” if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or “No” if not. Then in the next line, print the doubled number.
Sample Input: 1234567899
Sample Output: Yes\n2469135798

#include< stdio.h>
#include< stdbool.h>
#include< string.h>
char doubleit(char str[], int x);
bool ist(char str[], int length);
char ch = 32;//创建一个可能多出一位的最大的位数的数
bool ist(char str[],int last)//是否两者的数都互有涵盖
{
    int arry[10] = { 0 }, arrys[10] = { 0 };//记得桶排
    for (int x = 0; x <= last; x++)
        arry[str[x] - 48] = 1;
    ch=doubleit(str, last);
    for (int x = 0; x <= last; x++)
        arrys[str[x] - 48] = 1;
    if(ch!=32)
        arrys[ch - 48] = 1;
    int flag=1;
    for (int x = 0; x <= 9; x++)//比较同样记得可以用strcamp（字符串比较时）
        if (arry[x] != arrys[x])flag = 0;
    if (flag)
        return true;//0是false，非0都是true
    else
        return false;
}
char doubleit(char str[], int last)
{
    int x = 0,temp,lastone=0;//lastone用来记录进位,temp作为一个中间值
    for ( x = last; x>=0 ; x--)//开始乘法
    {
        temp = (str[x] - 48) * 2 ;//取模
        str[x] = temp%10 + 48+lastone;
        lastone = temp / 10;
        if (x == 0 && lastone != 0)
            ch = lastone + 48;
    }
    return ch;
}

int main()
{
    char str[22];
    int lastnumber;
    scanf("%s", str);//scanf会自动补'\0'
    lastnumber = strlen(str)-1;//strlen不包括'\0'
    if (ist(str, lastnumber))printf("Yes\n");
    else printf("No\n");
    if (ch != 32)
        printf("%c%s", ch, str);
    else
        printf("%s", str);
    return 0;
}

总结

大数据做运算用字符串来存放，注意的点
1.数据是“倒序存放”，这时就一定注意可能多出来一位，需要再构建一位
2.进位问题，有可能是1也有可能是2，所以用取模的方法就较为简单
3.若有对比问题，flag或者memset都要想到
4.同样可以建多数组，将计算结果直接存放在新数列