/*51nod 1770 数数字problem:给你a,b,d,n. 求n个a与b相乘后其中d的个数样例解释：3 3 9 103333333333*3=9999999999，里面有10个9solve:如果两者相乘小于10,那么和d比较进行判断.否则进行统计,会发现相乘到某一位时,它的进位会一直不变. 即n个a与b相乘后中间有一串数字是连续的,只要求到这串连续的开始位置即可.8 7 2 5--> 62216hhh-2016/09/03 12:13:07*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <vector>#include <math.h>#include <queue>#include <set>#include <map>#define lson  i<<1#define rson  i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define scanfi(a) scanf("%d",&a)#define scanfs(a) scanf("%s",a)#define scanfl(a) scanf("%I64d",&a)#define scanfd(a) scanf("%lf",&a)#define key_val ch[ch[root][1]][0]#define eps 1e-7#define inf 0x3f3f3f3f3f3f3f3fusing namespace std;const ll mod = 1e9+7;const int maxn = 20010;const double PI = acos(-1.0);int num[10];int main(){    int T,a,b,d,n;    scanfi(T);    while(T--)    {        clr(num,0);        scanfi(a),scanfi(b),scanfi(d),scanfi(n);        if(a * b < 10)        {            if(d == a*b)            {                printf("%d\n",n);            }            else                printf("0\n");        }        else        {            int preup = -1,up = 0;            for(int i = 1; i <= n; i++)            {                int leave = (a*b+up)%10;                up = (a*b+up)/10;                num[leave] ++ ;                if(up == preup)                {                    num[leave] += (n - i);                    num[up] ++ ;                    break;                }                if(i == n)                {                    num[up] ++ ;                }                preup = up;            }            printf("%d\n",num[d]);        }    }    return 0;}/*108 7 2 57 7 4 105 2 1 1381*/