Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

 I Love Big Numbers !

The Problem

A Japanese young girl went to a Science Fair at Tokyo. There she met with a Robot named Mico-12, which had AI (You must know about AI-Artificial Intelligence). The Japanese girl thought, she can do some fun with that Robot. She asked her, "Do you have any idea about maths ?"."Yes! I love mathematics", The Robot replied.

"Okey ! Then I am giving you a number, you have to find out the Factorial of that number. Then find the sum of the digits of your result!. Suppose the number is 5.You first calculate 5!=120, then find sum of the digits 1+2+0=3.Can you do it?"

"Yes. I can do!"Robot replied.

"Suppose the number is 100, what will be the result ?".At this point the Robot started thinking and calculating. After a few minutes the Robot head burned out and it cried out loudly "Time Limit Exceeds".

The girl laughed at the Robot and said "The sum is definitely 648".

"How can you tell that ?" Robot asked the girl. "Because I am an ACM World Finalist and I can solve the Big Number problems easily." Saying this, the girl closed her laptop computer and went away.

Now, your task is to help the Robot with the similar problem.

 

The Input

The input file will contain one or more test cases.

Each test case consists of one line containing an integers n (n<=1000).

The Output

For each test case, print one line containing the required number. This number will always fit into an integer, i.e. it will be less than 2^31-1.

Sample Input

5
60
100 

Sample Output

3
288
648

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作法 : 大數運算

#include<stdio.h>
int P[1001][2600] = {}, Ans[1001] = {};
int Build(int P[][2600]) {
    int i, j, last = 0;
    P[0][0] = 1, Ans[0] = 1;
    P[1][0] = 1, Ans[1] = 1;
    for(i = 2; i <= 1000; i++) {
        for(j = 0; j <= last; j++) {
            P[i][j] += P[i-1][j]*i;
            if(P[i][j] >= 10) {
                P[i][j+1] += P[i][j]/10;
                P[i][j] %= 10;
                if(j+1 > last)    last = j+1;
            }
            Ans[i] += P[i][j];
        }
    }
}
int main() {
    Build(P);
    int n;
    while(scanf("%d", &n) == 1)
        printf("%d\n", Ans[n]);
    return 0;
}