The Intersting Digit

The Intersting Digit

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

In this problem you will be given two decimal integer number N, M. You will have to find the last non-zero digit of the ^NP_M.This means no of permutations of N things taking M at a time.

输入

The input contains several lines of input. Each line of the input file contains two integers N (0 <= N<= 20000000), M (0 <= M <= N).

输出

For each line of the input you should output a single digit, which is the last non-zero digit of NPM. For example, if NPM is 720 then the last non-zero digit is 2. So in this case your output should be 2.

样例输入

10 1010 525 6

样例输出

842

来源

poj

这个题求的是c(n, m)的最后一个非零数， 但实际上求的是n！/(n-m)! 的最后一位非零数（汗-_-|||）详情可见网上的解释（poj1150）， 就是求2， 3， 5， 7， 9的个数来简化计算， 数论的题。。。

解释

#include <stdio.h>#include <math.h>int get2(int n){    if(n == 0)    {        return 0;    }    return n/2+get2(n/2);}int get5(int n){    if(n == 0)    {        return 0;    }    return n/5+get5(n/5);}int getxfromodd(int n, int x)      //求n！的 奇序列中找3， 7， 9的个数{    if(n == 0)    {        return 0;    }    return n/10 + (n%10 >= x) + getxfromodd(n/5, x);  //5, 15, 25, 35中也有3， 7， 9}int getx(int n, int x){    if(n == 0)    {        return 0;    }    return getxfromodd(n, x)+getx(n/2, x);    //先从奇数序列， 再从偶数序列中找3， 7， 9}int circle[4][4] = {   //2， 3， 7， 9的n次方循环节    6, 2, 4, 8,     //2    1, 3, 9, 7,     //3    1, 7, 9, 3,     //7    1, 9, 1, 9      //9};int main(){    /*for(int i = 0; i < 4; i++)    {        for(int j = 0; j < 4; j++)        {            printf("%d ", circle[i][j]);        }        printf("\n");    }*/    int n, m, num2, num3, num5, num7, num9, res;    while(scanf("%d%d", &n, &m) != EOF)    {        res = 1;   /*     num2 = get2(n)-(get2(m)+get2(n-m));        num5 = get5(n)-(get5(m)+get5(n-m));        num3 = getx(n, 3)-(getx(m, 3)+getx(n-m, 3));        num7 = getx(n, 7)-(getx(m, 7)+getx(n-m, 7));        num9 = getx(n, 9)-(getx(m, 9)+getx(n-m, 9));*/        num2 = get2(n) - get2(n-m);        num5 = get5(n) - get5(n-m);        num3 = getx(n, 3) - getx(n-m, 3);        num7 = getx(n, 7) - getx(n-m, 7);        num9 = getx(n, 9) - getx(n-m, 9);      //  printf("%d %d %d %d %d\n", num2, num3, num5, num7, num9);        if(num5 > num2)            //若5的个数大于2的， 结果为5        {            printf("5\n");            continue;        }        else                      //否则另算        {            if(num2 != num5)      //2^0为1但2^4为6， 需另算            {                res *= circle[0][(num2-num5)%4];                res %= 10;            }            res *= circle[1][num3%4];            res %= 10;            res *= circle[2][num7%4];            res %= 10;            res *= circle[3][num9%4];            res %= 10;            printf("%d\n", res);        }    }    return 0;}