UVa10220:I Love Big Numbers !

题目大意

给定正整数n，求n!的各位数字之和。

例：给定n为5，5!=120，答案是1+2+0=3

Solution

简单粗暴，直接求出n!然后计算各位数字之和。因为n最大可能为1000，所以需要采用高精度计算，具体实现方式与Uva623一致，可以参考我的另一篇博客。

AC-Code(C++)

Time:0ms

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <climits>
#include <ctime>

using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = 1000 + 10;


int number[maxn][maxn];
int len[maxn];

int getAns(int x){
    int ans = 0;
    for(int i=0;i<len[x];i++){
        int temp = number[x][i];
        while(temp > 0){
            ans += temp % 10;
            temp /= 10;
        }
    }
    return ans;
}

int main(int argc, const char * argv[]) {
    
//    freopen("input.txt", "r", stdin);
    
    int n;
    number[0][0] = 1;
    len[0] = 1;
    for(int i=1;i<=1000;i++){
        memcpy(&number[i][0], &number[i-1][0], len[i-1] * sizeof(int));
        len[i] = len[i-1];
        int carry = 0;
        for(int j=0;j<len[i-1];j++){
            number[i][j] *= i;
            number[i][j] += carry;
            carry  = number[i][j] / 10000; // stored 4 digit in each unit of number
            number[i][j] %= 10000;
        }
        if(carry > 0)
            number[i][len[i]++] = carry;
    }
    while(scanf("%d",&n)==1){
        printf("%d\n",getAns(n));
    }
    return 0;
}