一个特大数运算的程序：1+2+4+8+……+2^100=?

#include <stdio.h>

int per[500];
int total;

void n_num(int number)
{
    int i,j,temp,flag;
    per[0] = 1;
    total = 1;
    for (i=1,flag=0;i<=number;i++)
    {
        for (j=0,temp=0;j<total;j++)
        {
            temp = per[j]*2+flag;
            per[j] = temp%10;
            flag = temp/10;
        }
        while (flag)
        {
            per[total++] = flag%10;
            flag /= 10;
        }
    }   
}

void main()
{
    int i,number;
    printf("Input number:");
    scanf("%d",&number);
    n_num(number);
    printf("2^%d-1=",number);
    for (i=total-1;i>=0;i--)
    {
        if(i == 0)
            printf("%d",per[0]-1);
        else
            printf("%d",per[i]);
    }
    printf("/n共有%d位！/n",total);
}
////////////////////
2^101-1 = 2535301200456458802993406410751
共有31位！