The Huge One

ACCEPTED !!!  really HUGE but straightforward ...
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Question: The Huge One

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#include "iostream.h"
#include "string.h"

int divisible(char *num, int k) {
   int len = strlen(num), v, v0, v1, v2, last2, last3, sum, sum1, sum2, i;
   char bkp[1002];
   switch(k) {
   case 1:
        return 1;
   case 2:
        v0 = num[len - 1] - '0';
        if(v0 % 2 == 0) return 1;
        return 0;
   case 3:
        sum = 0;
        for(int i = 0; i < len; i++) sum += num[i] - '0';
        if(sum % 3 == 0) return 1;
        return 0;
   case 4:
        v0 = num[len - 1] - '0';
        v1 = num[len - 2] - '0';
        last2 = v1 * 10 + v0;
        if(last2 % 4 == 0) return 1;
        return 0;
   case 5:
        if((num[len - 1] - '0') % 5 == 0) return 1;
        return 0;
   case 6:
        v0 = num[len - 1] - '0';
        sum = 0;
        for(int i = 0; i < len; i++) sum += num[i] - '0';
        if(v0 % 2 == 0 && sum % 3 == 0) return 1;
        return 0;
   case 7:
        strcpy(bkp, num);
        i = 0;
        while(i < len - 1) {
            v1 = bkp[i] - '0';
            v0 = bkp[i + 1] - '0';
            v = v1 * 3 + v0;
            bkp[i] = v / 10 + '0';
            bkp[i + 1] = v % 10 + '0';
            if(bkp[i] == '0') i++;
        }
        if((bkp[i] - '0') % 7 == 0) return 1;
        return 0;
   case 8:
        v0 = num[len - 1] - '0';
        v1 = num[len - 2] - '0';
        v2 = num[len - 3] - '0';
        last3 = v2 * 100 + v1 * 10 + v0;
        if(last3 % 8 == 0) return 1;
        return 0;
   case 9:
        sum = 0;
        for(int i = 0; i < len; i++) sum += num[i] - '0';
        if(sum % 9 == 0) return 1;
        return 0;
   case 10:
        if((num[len - 1] - '0') == 0) return 1;
        return 0;
   case 11:
        sum1 = 0;
        sum2 = 0;
        for(int i = 0, j = 1; i < len || j < len; i += 2, j += 2) {
            if(i < len) sum1 += num[i] - '0';
            if(j < len) sum2 += num[j] - '0';
        }
        if((sum1 - sum2) % 11 == 0) return 1;
        return 0;
   case 12:
        sum = 0;
        for(int i = 0; i < len; i++) sum += num[i] - '0';
        v0 = num[len - 1] - '0';
        v1 = num[len - 2] - '0';
        last2 = v1 * 10 + v0;
        if(sum % 3 == 0 && last2 % 4 == 0) return 1;
        return 0;
   default:
        return 1;
   }
}

int main()
{
   int n;
   cin >> n;
   for(int t = 1; t <= n; t++) {
       char hugenum[1002];
       cin >> hugenum;
       int m;
       cin >> m;
       int j, k;
       for(j = 0; j < m; j++) {
           cin >> k;
           int r = divisible(hugenum, k);
           if(r == 0) break;
       }
       cout << hugenum;
       if(j == m) cout << " - Wonderful." << endl;
       else cout << " - Simple." << endl;
       while(m > ++j) cin >> k;
   }
   return 0;
}
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