POJ 3331 The Idiot of the Year Contest! 高精度阶乘

http://poj.org/problem?id=3331

The Idiot of the Year Contest!

Time Limit:2000MS		Memory Limit:65536K
Total Submissions:3072		Accepted:1590

Description

There is just one basic rule in the Idiot of the Year Contest (IYC)! The contestant picks a random digit between 0 and 9, computes the factorial of the day of the year he/she is born, and counts the how many times the digit picked appears in the factorial. The contestant with highest count is the Idiot of the Year! For example, if you are born on 5th of Mordad which is the 129th day of the year, and you pick the digit 6, your score will be the number of times the digit 6 appears in 129! (that is 1 × 2 × 3 × ... × 129).

The chief judge of IYC wants you to write a program to get an integer which is the day of the year a contestant is born on and a digit and report the number of times the digit appears in the factorial of the first number.

Input

The first line of the input contains a single integerTwhich is the number of test cases, followed byTlines each containing the data for a test case having two numbers. The first number is the day of the year a contestant is born and the second one is the digit he/she has picked.

Output

The output containsTlines, each having one integer which is the number of times the digit appears in the factorial of the first number.

Sample Input

2
5 2
7 0

Sample Output

1
2 

/* Author : yan
 * Question : POJ 3331 The Idiot of the Year Contest!
 * Date && Time : Monday, January 31 2011 08:07 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
#define MAX 1000
char mul1[MAX],mul2[4];
char ans[365][MAX];
void multiply(char* a,char* b,char* c)
{
    int i,j,ca,cb,* s;
    ca=strlen(a);
    cb=strlen(b);
    s=(int*)malloc(sizeof(int)*(ca+cb));
    for (i=0;i<ca+cb;i++)
        s[i]=0;
    for (i=0;i<ca;i++)
        for (j=0;j<cb;j++)
            s[i+j+1]+=(a[i]-'0')*(b[j]-'0');//i+j+1的目的就是为了防止最高位进位而产生错误
    for (i=ca+cb-1;i>=0;i--)
        if (s[i]>=10)
        {
            s[i-1]+=s[i]/10;
            s[i]%=10;
        }
    i=0;
    while (s[i]==0)
        i++;//去除前导0
       for (j=0;i<ca+cb;i++,j++)
           c[j]=s[i]+'0';
    c[j]='/0';//将结果存储到字符数组
    free(s);
}
int main()
{
	//freopen("input","r",stdin);
	int t,i;
	int day,digit,tmp,res;
	char cache[4];
	ans[0][0]='1',ans[0][1]='/0';
	for(i=1;i<366;i++)
	{
		sprintf(cache,"%d",i);
		multiply(ans[i-1],cache,ans[i]);
	}
	scanf("%d",&t);
	while(t--)
	{
		res=0;
		scanf("%d %d",&day,&digit);
		tmp=strlen(ans[day]);
		for(i=0;i<tmp;i++)
		{
			if(ans[day][i]-'0'==digit) res++;
		}
		printf("%d/n",res);
	}
	return 0;
}