题目

Description

问题描述：统计n阶乘中数字p的个数

Input

多组数据，每组数据只有一行，为两个正整数n,p (n<=800,0<=p<=9)

Output

输出为两行，一行为n!，另一行为n! 中数字p的个数

Sample Input

10 8

Sample Output

3628800
2
题解

第一次把 800 看成 8000 ,优化高精度优化了好久
代码

/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/
 
かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
 
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
using namespace std;
 
#define REP(n) for(int o=0;o<n;o++)

class bigNumber {
public:
    int num[10001];
    bigNumber() {
        init();
    }
    void init() {
        memset(num,0,sizeof(num));
    }
    bigNumber operator = (const bigNumber& rhs) {
        init();
        REP(rhs.num[0] + 1)num[o] = rhs.num[o];
        return *this;
    }
    bigNumber operator = (long long rhs) {
        init();
        int i = 1;
        while(rhs) {
            num[i] = rhs % 10;
            rhs /= 10;
            i++;
        }
        num[0] = i - 1;
        return *this;
    }
    bool operator < (const bigNumber rhs)const {
        if(num[0] != rhs.num[0])return (num[0]<rhs.num[0]);
        REP(num[0]) {
            int temp = num[0] - o;
            if(num[temp] != rhs.num[temp])return (num[temp]<rhs.num[temp]);
        }
        return 0;
    }
    bool operator > (const bigNumber rhs)const {
        if(num[0] != rhs.num[0])return (num[0]>rhs.num[0]);
        REP(num[0]) {
            int temp = num[0] - o;
            if(num[temp] != rhs.num[temp])return (num[temp]>rhs.num[temp]);
        }
        return 0;
    }
    bool operator == (const bigNumber rhs)const {
        return !(*this>rhs || *this<rhs);
    }
    bool operator <= (const bigNumber rhs)const {
        return *this<rhs || *this == rhs;
    }
    bigNumber operator + (const bigNumber rhs)const {
        bigNumber temp;
        int len;
        len = num[0]>rhs.num[0] ? num[0] : rhs.num[0];
        len++;
        REP(len) {
            temp.num[o + 1] += num[o + 1] + rhs.num[o + 1];
            temp.num[o + 2] += temp.num[o + 1] / 10;
            temp.num[o + 1] %= 10;
        }
        REP(len) {
            if(temp.num[len - o] != 0) {
                temp.num[0] = len - o;
                break;
            }
        }
        return temp;
    }
    bigNumber operator + (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 + temp2;
    }
    bigNumber operator * (const bigNumber rhs)const {
        bigNumber temp;
        int len;
        len = num[0] + rhs.num[0];
        //len++;
        for(int i = 1;i <= num[0];i++) {
            for(int j = 1;j <= rhs.num[0];j++) {
                temp.num[i + j - 1] += num[i] * rhs.num[j];
                temp.num[i + j] += temp.num[i + j - 1] / 10;
                temp.num[i + j - 1] %= 10;
            }
        }
        REP(len) {
            if(temp.num[len - o] != 0) {
                temp.num[0] = len - o;
                break;
            }
        }
        return temp;
    }
    bigNumber operator * (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 * temp2;
    }
    bigNumber operator - (const bigNumber rhs)const {
        bigNumber temp,a,b;
        temp = max(*this,rhs);
        b = min(*this,rhs);
        a = temp;
        temp = 0;
        int len = a.num[0];
        REP(len) {
            temp.num[o + 1] += 10 + a.num[o + 1] - b.num[o + 1];
            temp.num[o + 2]--;
            temp.num[o + 2] += temp.num[o + 1] / 10;
            temp.num[o + 1] %= 10;
        }
        REP(len) {
            if(temp.num[len - o] != 0) {
                temp.num[0] = len - o;
                break;
            }
        }
        return temp;
    }
    bigNumber operator - (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 - temp2;
    }
    bigNumber operator / (const bigNumber rhs)const {
        bigNumber a;
        int it = num[0];
 
        bigNumber d;
        bigNumber c;
 
        while(it>0) {
            a = (d * 10) + num[it];
            c = c * 10;
 
            int t;
            REP(9) {
                if(a < rhs * (o + 1)) {
                    t = o;
                    break;
                }
                t = 9;
            }
            c = c + t;
            d = a - rhs*t;
            it--;
        }
        return c;
    }
    bigNumber operator / (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 / temp2;
    }
    bigNumber operator % (const bigNumber rhs)const {
        bigNumber a;
        int it = num[0];
 
        bigNumber d;
        bigNumber c;
 
 
 
        while(it>0) {
 
            a = (d * 10) + num[it];
            c = c * 10;
 
            int t;
            REP(9) {
                if(a < rhs * (o + 1)) {
                    t = o;
                    break;
                }
                t = 9;
            }
            c = c + t;
            d = a - rhs*t;
            it--;
        }
        return d;
    }
    bigNumber operator % (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 % temp2;
    }
    void p() {
        if(num[0] == 0)
            printf("0");
        REP(num[0]) {
            printf("%d",num[num[0] - o]);
        }
    }
};
 
bigNumber ans;
 
bool Do() {
    int n,p;
    if(scanf("%d%d",&n,&p) == EOF)
        return false;
     
    ans = 1;
    int cnt = 0;
 
    for(int i = 2;i <= n;i++) {
        ans = ans * i;
    }
 
    REP(ans.num[0]) {
        printf("%d",ans.num[ans.num[0] - o]);
        if(ans.num[ans.num[0] - o] == p)cnt++;
    }
 
    printf("\n%d\n",cnt);
    return true;
}
 
int main() {
    while(Do());
    return 0;
}
AOJ 787.阶乘中找数
